Look at 5, 11, 17, 23, … — every new number comes from adding 6 to the one before it. Look at 6, 12, 24, 48, … — every new number comes from doubling the one before it. Whenever numbers follow a clear rule like this, so that you can always say what comes next, the list is a sequence. More formally, a sequence is a function whose domain is the set N of natural numbers, while its values may be real or complex numbers.
The numbers in the list are its terms: the first term is a₁, the second a₂, and the nth term an is called the general term. If you know the general term, you know the whole sequence — just put n = 1, 2, 3, … in turn. Example 1 of the book does exactly this: for an = 3n + 1 the first four terms are 4, 7, 10, 13, and for an = 3n² − 3 they are 0, 9, 24, 45.
📌 Book note: a sequence given by its terms lets you find its nth term, and a sequence given by its nth term lets you find its terms — the two directions are the same skill seen from opposite sides. Sequences are also called progressions, and the book sorts them into three families: arithmetic, geometric and harmonic.
A sequence with a fixed, finite number of terms is finite — for example 2, 5, 8, 11, 14, 17, 20, 23 has 8 terms and then stops. A sequence that goes on without end is infinite, written with dots, like 3, 10, 17, 24, … — or, better, with its general term shown, 3, 10, 17, 24, …, 7n − 4, …, so the reader can see how each term was made.
جب اعداد کسی واضح اصول کے مطابق ترتیب سے آئیں اور ہر اگلا عدد معلوم کیا جا سکے تو اس فہرست کو متوالیہ کہتے ہیں۔ مثلاً 5، 11، 17، 23، … میں ہر عدد پچھلے میں 6 جمع کرنے سے بنتا ہے، اور 6، 12، 24، 48، … میں ہر عدد پچھلے کا دوگنا ہے۔
متوالیے کے اعداد اس کی رقمیں (terms) کہلاتے ہیں: پہلی رقم a₁، دوسری a₂، اور n ویں رقم an کو عمومی رقم (general term) کہتے ہیں۔ عمومی رقم معلوم ہو تو n = 1، 2، 3، … رکھ کر پوری متوالیہ بنائی جا سکتی ہے، جیسے an = 3n + 1 سے 4، 7، 10، 13 بنتے ہیں۔
📌 کتابی نوٹ: رقمیں دی ہوں تو عمومی رقم نکالی جا سکتی ہے، اور عمومی رقم دی ہو تو رقمیں نکالی جا سکتی ہیں۔ جس متوالیے کی رقموں کی تعداد مقرر ہو وہ محدود اور جو کبھی ختم نہ ہو وہ لامحدود کہلاتی ہے۔
An arithmetic progression is a sequence in which every term after the first is got by adding one fixed number to the term before it. That fixed number is the common difference d, and it is the same all the way through: d = a₂ − a₁ = a₃ − a₂ = … . So 1, 3, 5, 7, … is an A.P. with d = 2, and 54, 51, 48, … is an A.P. with d = −3 — a negative difference just means the sequence falls instead of rising.
Writing the terms out, an A.P. with first term a₁ looks like a₁, a₁ + d, a₁ + 2d, …, a₁ + (n − 1)d, and that last expression is the famous nth term formula: an = a₁ + (n − 1)d. Count the steps, not the terms: to reach the nth term you take (n − 1) steps of size d. That one observation answers most A.P. questions — Example 2 (a₁ = 2, d = −3) gives an = 5 − 3n and a₁₁ = −28, and Example 4 turns an = 59 with a₁ = 3, d = 7 into n = 9.
📌 Book notes: (i) three numbers a, b, c are in A.P. exactly when 2b = a + c; (ii) any term except the first and last equals half the sum of two terms equally far from it on either side; (iii) if a₁ is not known, the nth term can be written from any known term as an = am + (n − m)d — notice the subscript of the known term and the coefficient of d add up to n. Middle terms: if the number of terms is even there are two middle terms, the (n/2)th and (n/2 + 1)th — in 1, 3, 5, 7, 9, 11 they are 5 and 7; if it is odd there is one, the ((n + 1)/2)th — in 1, 3, 5, 7, 9, 11, 13 it is 7.
Selection of terms is a bookkeeping trick for word problems: three consecutive terms are best written a − d, a, a + d (their sum is just 3a, no d left over); four as a − 3d, a − d, a + d, a + 3d; and the last four, if ℓ is the last term, as ℓ − 3d, ℓ − 2d, ℓ − d, ℓ. Finally, adding or subtracting the same number to every term, multiplying or dividing every term by the same non-zero number, and adding or subtracting two A.P.s term by term all leave you with an A.P. again.
حسابی متوالیہ وہ متوالیہ ہے جس میں پہلی رقم کے بعد ہر رقم پچھلی رقم میں ایک مقرر عدد جمع کرنے سے بنے۔ اس مقرر عدد کو مشترکہ فرق d کہتے ہیں، یعنی d = a₂ − a₁ = a₃ − a₂ = …۔ مثلاً 1، 3، 5، 7، … میں d = 2 ہے اور 54، 51، 48، … میں d = −3 ہے۔
عمومی فارمولا an = a₁ + (n − 1)d ہے۔ یاد رکھیں: n ویں رقم تک پہنچنے کے لیے (n − 1) قدم چلنا ہوتا ہے، اس لیے d کو (n − 1) سے ضرب دیتے ہیں، n سے نہیں۔
📌 کتابی نوٹ: تین اعداد a، b، c حسابی متوالیہ میں ہوں تو 2b = a + c ہوتا ہے۔ درمیانی رقموں کے لیے: رقموں کی تعداد جفت ہو تو دو درمیانی رقمیں ہوتی ہیں، طاق ہو تو ایک۔ سوالوں میں تین مسلسل رقمیں a − d، a، a + d لکھنا آسان رہتا ہے کیونکہ ان کا مجموعہ صرف 3a بنتا ہے۔
A number A is the arithmetic mean between a and b when a, A, b are in A.P. — in other words, when A sits exactly halfway between them, so that the step from a to A equals the step from A to b: A − a = b − A. Solving gives the familiar result A = (a + b)/2, the ordinary average of the two numbers. Nothing more mysterious than that hides behind the name.
📌 Book note: if A₁, A₂, …, An are n arithmetic means between a and b, then a, A₁, A₂, …, An, b is one long A.P. with n + 2 terms. That single fact does all the work in insertion problems: the common difference is d = (b − a)/(n + 1), because there are n + 1 steps from a to b. Example 6 inserts three means between √2 and 3√2: the five-term A.P. has d = 1/√2, so the means are 3/√2, 2√2 and 5/√2.
Watch the counting in these questions — students usually miscount the number of terms. Between a and b with n means inserted, the total number of terms is n + 2, the last term b is the (n + 2)th term, and the step count from a to b is n + 1. Get that right and the rest is plain substitution.
اگر a، A، b حسابی متوالیہ میں ہوں تو A کو a اور b کے درمیان حسابی اوسط کہتے ہیں۔ چونکہ A − a = b − A ہوتا ہے، اس لیے A = (a + b)/2 نکلتا ہے، یعنی دونوں اعداد کا عام اوسط۔
📌 کتابی نوٹ: اگر a اور b کے درمیان n حسابی اوسطیں A₁، A₂، …، An داخل کی جائیں تو a، A₁، …، An، b ایک حسابی متوالیہ بن جاتی ہے جس کی کل رقموں کی تعداد n + 2 اور مشترکہ فرق d = (b − a)/(n + 1) ہوتا ہے، کیونکہ a سے b تک n + 1 قدم ہوتے ہیں۔
A series is what you get when you add the terms of a sequence instead of just listing them: 1 + 2 + 3 + … + n is a finite series, and the sum of its first n terms is written Sn. The running totals S₁ = a₁, S₂ = a₁ + a₂, S₃ = a₁ + a₂ + a₃, … are called partial sums. The distinction is small but it matters: a sequence is a list, a series is a sum.
The book derives the arithmetic sum the way Gauss is said to have done it as a child — write the sum forwards, write it backwards underneath, and add the two lines term by term. Every pair adds to (a₁ + an), there are n pairs, and two copies of the sum make n(a₁ + an), so Sn = (n/2)(a₁ + an). Since an = a₁ + (n − 1)d, the same formula can be written Sn = (n/2)[2a₁ + (n − 1)d] — use the first form when you know the last term, the second when you know d.
📌 Key concept (book box): both forms are the same formula, not two different ones, and Example 7 checks both on 1 + 2 + … + 100: each gives S₁₀₀ = 50 × 101 = 5050. Two habits save marks here: when a sum equation turns into a quadratic in n, throw away the negative root (a count of terms cannot be negative — see Example 10, where n = 11, −4 leaves n = 11), and when a question gives you Sn together with a₁ and an, reach for the (a₁ + an) form first, as in Example 11, where it finds n = 13 before d = 8 is even needed.
متوالیے کی رقموں کو جمع کرنے سے سلسلہ بنتا ہے۔ پہلی n رقموں کے مجموعے کو Sn لکھتے ہیں، اور S₁، S₂، S₃، … کو جزوی مجموعے کہتے ہیں۔
حسابی سلسلے کا فارمولا مجموعے کو آگے اور پیچھے دونوں طرف لکھ کر نکالا جاتا ہے: ہر جوڑا (a₁ + an) بنتا ہے اور n جوڑے ہوتے ہیں، اس لیے Sn = (n/2)(a₁ + an) اور Sn = (n/2)[2a₁ + (n − 1)d]۔ آخری رقم معلوم ہو تو پہلا فارمولا، d معلوم ہو تو دوسرا استعمال کریں۔
📌 اہم بات: مجموعے کی مساوات سے n کی دو قدریں نکلیں تو منفی قدر چھوڑ دیں، کیونکہ رقموں کی تعداد منفی نہیں ہو سکتی۔
Where an A.P. adds the same number each time, a geometric progression multiplies by the same non-zero number each time. That number is the common ratio r, found by dividing any term by the one before it: r = an/an−1. So 3, 12, 48, … has r = 4, and −6, −3, −3/2, … has r = 1/2. A negative ratio makes the signs alternate, and a ratio between −1 and 1 makes the terms shrink towards zero — both are normal behaviour, not errors.
Following the multiplications gives the nth term: an = a₁rn−1. The power is n − 1 because the first term has not been multiplied at all yet; students who write rn here are the ones who lose the mark. Example 12 (a₁ = −3, r = −2) gives a₈ = (−3)(−2)⁷ = 384, and Example 14 works backwards from a₄ = 108, r = 3 to a₁ = 4 and then a₁₀ = 78,732.
📌 Book properties: (i) multiplying or dividing every term by the same non-zero number leaves a G.P.; (ii) reciprocals of a G.P. form a G.P.; (iii) raising every term to the same power leaves a G.P.; (iv) three numbers a, b, c are in G.P. exactly when b² = ac; (v) logs of a positive G.P. form an A.P., and vice versa; (vi) term-by-term multiplication or division of two G.P.s gives a G.P. again. Keep b² = ac beside 2b = a + c in your mind — the two tests are twins, one for each family.
ہندسی متوالیہ میں ہر رقم پچھلی رقم کو ایک ہی غیر صفر عدد سے ضرب دے کر بنتی ہے۔ اس عدد کو مشترکہ نسبت r کہتے ہیں، جو کسی بھی رقم کو اس سے پچھلی رقم پر تقسیم کرنے سے ملتی ہے: r = an/an−1۔
عمومی فارمولا an = a₁rn−1 ہے۔ طاقت n − 1 اس لیے ہے کہ پہلی رقم پر ابھی کوئی ضرب نہیں لگی۔ منفی نسبت سے علامتیں باری باری بدلتی ہیں، یہ کوئی غلطی نہیں۔
📌 کتابی خصوصیات: تین اعداد a، b، c ہندسی متوالیہ میں ہوں تو b² = ac ہوتا ہے؛ ہر رقم کو ایک ہی عدد سے ضرب یا تقسیم کرنے، یا معکوس لینے سے پھر ہندسی متوالیہ ہی بنتی ہے۔
A number G is the geometric mean between a and b when a, G, b are in G.P. That means the ratio from a to G equals the ratio from G to b: G/a = b/G, and cross-multiplying gives G² = ab, so G = ±√(ab). There are two square roots, so there are two geometric means in general — the positive one is the one people usually mean when they say "the" geometric mean of two positive numbers.
📌 Book note: if G₁, G₂, …, Gn are n geometric means between a and b, then a, G₁, …, Gn, b is a G.P. with n + 2 terms, and the ratio comes from the last term: b = a·rn+1. Example 17 inserts three means between 2 and 1/2: r⁴ = 1/4, and the book honestly follows all four roots — r = ±1/√2 and r = ±i/√2 — giving four different triples of means, with a note that real values of r are the ones usually taken. That note is worth remembering: a question that says "real geometric means" (as Exercise 6.6 Q2 does) is telling you to drop the complex roots.
The most useful comparison in the whole unit is the one Exercise 6.6 Q3 asks you to prove: for two distinct positive numbers, the geometric mean is less than the arithmetic mean. It follows from (√x − √y)² > 0, and it is the reason averages of growth rates and ratios are usually geometric, not arithmetic.
اگر a، G، b ہندسی متوالیہ میں ہوں تو G کو a اور b کے درمیان ہندسی اوسط کہتے ہیں۔ اس صورت میں G/a = b/G ہوتا ہے، یعنی G² = ab اور G = ±√(ab)۔ مثبت اعداد کے لیے عام طور پر مثبت جزر ہی مراد لیا جاتا ہے۔
📌 کتابی نوٹ: اگر a اور b کے درمیان n ہندسی اوسطیں داخل کی جائیں تو کل n + 2 رقموں کی ہندسی متوالیہ بنتی ہے اور نسبت b = a·rn+1 سے ملتی ہے۔ مثال 17 میں 2 اور 1/2 کے درمیان تین اوسطوں کے لیے r کی چار قدریں نکلتی ہیں، جن میں مخلوط قدریں بھی شامل ہیں؛ عام طور پر صرف حقیقی قدریں لی جاتی ہیں۔
The book opens this section with an e-mail that spreads to three friends, who each send it to three more, and so on: 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 = 3280 readers by Sunday. The sum of a geometric sequence is a geometric series, and its formula comes from a trick very like the A.P. pairing: write Sn, multiply the whole line by r, and subtract. Almost everything cancels, leaving Sn − rSn = a₁ − a₁rn, so Sn = a₁(1 − rn)/(1 − r), for r ≠ 1.
📌 Book note: if r = 1, every term equals a₁ and the sum is simply Sn = na₁ — the formula above would divide by zero, which is why the condition r ≠ 1 is printed beside it. Example 18 shows the other skill this section teaches: recognising a₁ and r inside a general term. From an = (−3)(2/5)n, rewrite it as (−6/5)(2/5)n−1 and read off a₁ = −6/5, r = 2/5, giving Sn = (−2)[1 − (2/5)n].
When |r| < 1, the terms shrink to nothing as n grows, and the same formula gives the sum to infinity: S∞ = a₁/(1 − r). That is the result Exercise 6.11 Q7 uses for the toll booth: 100 vehicles a minute, each minute 80% of the last, can never pass more than 100/(1 − 0.8) = 500 vehicles in total, however long you wait.
ہندسی متوالیہ کی رقموں کے مجموعے کو ہندسی سلسلہ کہتے ہیں۔ کتاب کی مثال میں ایک پیغام تین دوستوں کو، پھر ہر دوست تین کو بھیجتا ہے: 1 + 3 + 9 + … + 2187 = 3280۔ فارمولا Sn کو r سے ضرب دے کر تفریق کرنے سے نکلتا ہے: Sn = a₁(1 − rn)/(1 − r)، بشرطیکہ r ≠ 1۔
📌 کتابی نوٹ: اگر r = 1 ہو تو ہر رقم a₁ کے برابر ہوتی ہے اور Sn = na₁ ہوتا ہے۔ اور جب |r| < 1 ہو تو لامحدود مجموعہ S∞ = a₁/(1 − r) ہوتا ہے۔
Multiply an A.P. term by term with a G.P. and you get an arithmetico-geometric sequence: with A.P. a, a + d, a + 2d, … and G.P. b, br, br², …, the product sequence is ab, (a + d)br, (a + 2d)br², …, {a + (n − 1)d}brn−1. Its nth term is simply the product of the two nth terms — that is the whole definition, and it is also the quickest way to spot an A.G.P. in a question: the coefficients form an A.P. while the powers form a G.P.
The sum uses the same multiply-and-subtract trick as the G.P. sum, and the book's result is Sn = ab/(1 − r) + dbr/(1 − r)² − dbrn/(1 − r)² − [a + (n − 1)d]brn/(1 − r), for r ≠ 1. It looks heavy, but in practice you rarely quote it whole: Examples 19–21 work the trick out by hand each time, which is both safer and easier to follow. Example 19 (2·1 + 3·2 + 4·4 + 5·8 + …) collapses beautifully to Sn = n·2n — check it for n = 1, 2, 3 and you will believe it.
📌 Sum to infinity: when |r| < 1, the terms involving rn and nrn die away as n → ∞, leaving S∞ = ab/(1 − r) + dbr/(1 − r)². Example 21 applies the same method to 1 + 2x + 3x² + 4x³ + … (x ≠ 1), giving Sn = [1 − (n + 1)xn + nxn+1]/(1 − x)² and, for |x| < 1, S∞ = 1/(1 − x)².
حسابی متوالیہ کی رقموں کو ہندسی متوالیہ کی متعلقہ رقموں سے ضرب دینے سے حسابی ہندسی متوالیہ بنتی ہے: ab، (a + d)br، (a + 2d)br²، …۔ اس کی n ویں رقم دونوں متوالیوں کی n ویں رقموں کی ضرب ہوتی ہے۔
مجموعہ نکالنے کے لیے وہی طریقہ استعمال ہوتا ہے: Sn کو r سے ضرب دے کر تفریق کریں۔ کتاب کا فارمولا Sn = ab/(1 − r) + dbr/(1 − r)² − dbrn/(1 − r)² − [a + (n − 1)d]brn/(1 − r) ہے۔ اور جب |r| < 1 ہو تو لامحدود مجموعہ S∞ = ab/(1 − r) + dbr/(1 − r)² رہ جاتا ہے۔
A sequence is a harmonic progression when the reciprocals of its terms form an arithmetic progression. There is no direct formula for the nth term of an H.P. — you always work through the reciprocal A.P. instead. For example, 1, 1/3, 1/5, 1/7 is harmonic because 1, 3, 5, 7 is arithmetic, and Example 22 finds the nth term of 1/2, 1/5, 1/8, … by first finding the A.P. 2, 5, 8, … with general term 3n − 1, then flipping it back: the H.P. term is 1/(3n − 1), and the 8th term is 1/23.
One warning the book prints in plain words: the reciprocal of zero is not defined, so zero can never be a term of a harmonic sequence. If a calculation hands you a zero term in an H.P., something has gone wrong upstream. The general form of an H.P. is 1/a₁, 1/(a₁ + d), 1/(a₁ + 2d), …, 1/(a₁ + (n − 1)d), where a₁ and d belong to the reciprocal A.P.
A number H is the harmonic mean between a and b when a, H, b are in H.P. — that is, when 1/a, 1/H, 1/b are in A.P. Taking the middle term as the average of its neighbours gives 1/H = (1/a + 1/b)/2, so H = 2ab/(a + b). For 3 and 7, that is 2 × 21/10 = 21/5. And to insert n harmonic means between a and b, the book's shortcut is: insert n arithmetic means between 1/a and 1/b, then take reciprocals of the results — Example 24 does this for 1/5 and 1/17, getting 1/8, 1/11, 1/14.
جب کسی متوالیہ کی رقموں کے معکوس حسابی متوالیہ بنائیں تو وہ متوالیہ توافقی متوالیہ کہلاتی ہے۔ توافقی متوالیہ کی n ویں رقم کا کوئی براہ راست فارمولا نہیں ہوتا — ہمیشہ معکوس والی حسابی متوالیہ سے کام لیا جاتا ہے، پھر جواب کا معکوس لے لیا جاتا ہے۔
یاد رکھیں: صفر کا معکوس متعین نہیں، اس لیے صفر توافقی متوالیہ کی رقم کبھی نہیں ہو سکتا۔
اگر a، H، b توافقی متوالیہ میں ہوں تو H کو a اور b کے درمیان توافقی اوسط کہتے ہیں، اور H = 2ab/(a + b) ہوتا ہے۔ n توافقی اوسطیں داخل کرنے کا طریقہ یہ ہے کہ پہلے 1/a اور 1/b کے درمیان n حسابی اوسطیں نکالیں، پھر ان کے معکوس لے لیں۔
The Greek letter Σ (sigma) is shorthand for "add them up". The sum am + am+1 + … + an is written Σk=mn ak, where k is the index of summation and m and n are the lower and upper limits. So 1 + 3 + 5 + … to n terms is Σk=1n (2k − 1), and the three famous sums are Σk, Σk² and Σk³ for k = 1 to n.
All three come from one telescoping trick. Expand [km − (k − 1)m] and add from k = 1 to n: every middle term appears once positive and once negative, so the whole sum collapses to nm. With m = 2, the identity k² − (k − 1)² = 2k − 1, summed, gives n² = 2Σk − n, and therefore Σk = n(n + 1)/2. The book states the other two without full working — "similarly, we can prove easily" — and they are Σk² = n(n + 1)(2n + 1)/6 and Σk³ = [n(n + 1)/2]². Notice the last one is just the square of the first: the sum of cubes is the square of the sum of the numbers themselves.
📌 Properties of summation (book box): Σ(ak + bk) = Σak + Σbk, and Σ(αak) = αΣak — a sum splits over addition and a constant factor pulls out front. With those two rules and the three formulas, almost every series in Exercise 6.10 falls apart: write the nth term, split it into powers of k, and sum each piece separately. Example 25 does this for 1³ + 3³ + 5³ + …: the term is (2k − 1)³ = 8k³ − 12k² + 6k − 1, and the pieces add up to n²(2n² − 1).
یونانی حرف Σ (سگما) کا مطلب ہے "جمع کریں"۔ am + am+1 + … + an کو Σk=mn ak لکھتے ہیں، جہاں k جمع کا اشاریہ اور m و n نچلی و اوپری حدیں ہیں۔
تین مشہور فارمولے یہ ہیں: Σk = n(n + 1)/2، Σk² = n(n + 1)(2n + 1)/6 اور Σk³ = [n(n + 1)/2]²۔ آخری فارمولا دراصل پہلے فارمولے کا مربع ہے۔ یہ سب ایک ہی ترکیب سے نکلتے ہیں: [km − (k − 1)m] کا مجموعہ لینے سے درمیانی رقمیں ختم ہو جاتی ہیں اور صرف nm بچتا ہے۔
📌 جمع کی خصوصیات: Σ(ak + bk) = Σak + Σbk اور Σ(αak) = αΣak۔ یعنی مجموعہ جمع پر تقسیم ہو جاتا ہے اور مستقل عدد باہر نکالا جا سکتا ہے۔
This last section is where the unit earns its keep: the same formulas, now wearing everyday clothes. Cars arriving at a toll booth at 4 every 5 minutes form an A.P., and after an hour the 12th term is 48 cars (Example 27). Simple interest grows by the same amount every year, so it is an A.P. too — Tayyab's Rs. 50,000 at 8% pays Rs. 4,000, Rs. 8,000, … and in the 11th year Rs. 44,000 (Example 28). Compound interest multiplies by the same factor every year, so it is a G.P.: Amna's Rs. 200,000 at 5% becomes 200,000 × (1.05)¹⁰ = Rs. 325,778.92 after ten years (Example 29).
The section then stretches further: a web designer's 12 grid columns, widening by 10px from 50px, total 1,260px (Example 30), and a motor-vehicle lease starting at Rs. 15,000 a month and rising Rs. 500 a month — 24th payment Rs. 26,500, total Rs. 498,000, which busts a Rs. 400,000 budget, leaving 20 months as the most the company can afford (Example 31, where the quadratic n² + 59n − 1600 ≤ 0 gives n ≈ 20.2, so n = 20).
The skill being tested in Exercise 6.11 is not new algebra but translation: which family is this — adding the same, multiplying by the same, or a sum? Once you have named the family and its a₁, d or r, the rest is a formula you already know. Read the question twice, underline the numbers that repeat, and ask what stays the same from step to step.
اس آخری حصے میں وہی فارمولے روزمرہ مسائل میں استعمال ہوتے ہیں۔ ٹول بوتھ پر ہر 5 منٹ بعد 4 گاڑیاں آنا حسابی متوالیہ ہے؛ ایک گھنٹے بعد 12 ویں رقم 48 بنتی ہے۔ سادہ سود ہر سال برابر بڑھتا ہے، اس لیے یہ بھی حسابی متوالیہ ہے، جبکہ مرکب سود ہر سال ایک ہی نسبت سے بڑھتا ہے، اس لیے یہ ہندسی متوالیہ ہے۔
مشق 6.11 میں اصل مہارت ترجمہ ہے: پہچانو کہ مسئلہ کس خاندان سے ہے — برابر جمع ہو رہا ہے، برابر ضرب لگ رہی ہے، یا مجموعہ مانگا جا رہا ہے؟ خاندان پہچان لیا تو باقی کام وہی فارمولا ہے جو آپ پہلے سے جانتے ہیں۔
Below are all questions of Exercises 6.1 to 6.11 from Unit 6 of the official book (printed pages 89, 93–94, 95, 98–99, 101–102, 103–104, 105, 109–110, 113–114, 117, 121–122), each solved step by step. The book has no multiple-choice section, no review exercise and no inquisitive/numerical section in this unit, so none is invented here. Where a step is derived rather than quoted from a book passage, or where the book's wording needs a note, the answer says so. Counting rule: one numbered question = one answer block; its parts (i), (ii), … are solved inside that block. That gives 4 + 23 + 7 + 19 + 15 + 8 + 6 + 11 + 18 + 4 + 15 = 130 blocks.
(i) The difference is steady: 16 − 12 = 4 and 20 − 16 = 4, so d = 4. Keep adding 4: 20 + 4 = 24, 24 + 4 = 28, 28 + 4 = 32, 32 + 4 = 36. (ii) Here 1 − 3 = −2 and −1 − 1 = −2, so d = −2. Keep adding −2: −1 − 2 = −3, −3 − 2 = −5, −5 − 2 = −7, −7 − 2 = −9.
(i) Put n = 1, 2, 3: a₁ = 3 + 5 = 8, a₂ = 6 + 5 = 11, a₃ = 9 + 5 = 14. (ii) a₁ = 3 is given. a₂ = 4(3) − 7 = 5; a₃ = 4(5) − 7 = 13. (iii) a₁ = (1 − 3)(1 + 1) = (−2)(2) = −4; a₂ = (2 − 3)(2 + 1) = (−1)(3) = −3; a₃ = (3 − 3)(3 + 1) = 0 × 4 = 0. (iv) a₁ = −1. a₂ = 3/(−1 + 2) = 3/1 = 3; a₃ = 3/(3 + 2) = 3/5. (v) a₁ = 8 − 20/4 = 8 − 5 = 3; a₂ = 8 − 20/5 = 8 − 4 = 4; a₃ = 8 − 20/6 = 8 − 10/3 = 14/3. (vi) a₁ = 1. a₂ = (3 + 2)² = 25; a₃ = (3 × 25 + 2)² = 77² = 5929. (vii) (−2n)² = 4n², so the terms are 4 × 1 = 4, 4 × 4 = 16, 4 × 9 = 36. (viii) a₁ = (−1) × 7 = −7; a₂ = (+1) × 28 = 28; a₃ = (−1) × 63 = −63.
Put n = 15 in the formula: T₁₅ = 15 × 16/2 = 240/2 = 120. For n = 5, draw dots in rows of 1, 2, 3, 4 and 5 — a triangle with 5 dots along each side — and count: 1 + 2 + 3 + 4 + 5 = 15 dots, which is exactly T₅ = 5 × 6/2 = 15, as it must be.
(i) The terms are 1², 2², 3², …, so an = n². (ii) The nth term is the sum 1 + 2 + … + n, so an = n(n + 1)/2. (iii) Each term pairs the same-indexed terms, so an = anbn. (iv) The number in front matches the position and the power of x matches it too, so an = nxn. (v) This is an A.P. with first term a₁ and difference d, so an = a₁ + (n − 1)d. (vi) This is a G.P. with first term a₁ and ratio r, so an = a₁rn−1. (vii) The numerator is n times an and the denominator is bn + cn, so an = nan/(bn + cn). (viii) The denominators form the A.P. a₁, a₁ + d, a₁ + 2d, …, so an = 1/(a₁ + (n − 1)d).
(i) d = 16 − 9 = 7 (check: 23 − 16 = 7). Next terms: 23 + 7 = 30, 30 + 7 = 37. (ii) d = (5 + √2) − 5 = √2 (check: (5 + 2√2) − (5 + √2) = √2). Next terms: 5 + 2√2 + √2 = 5 + 3√2, then 5 + 4√2.
(i) Start at 2 and add 13 each time: 2, 15, 28. (ii) Start at 12 and add −13 each time: 12, −1, −14.
Replace n by n + 1: an+1 = 4 + 3(n + 1) = 4 + 3n + 3 = 7 + 3n. Replace n by 2n: a2n = 4 + 3(2n) = 4 + 6n. The trap here is forgetting to multiply out the bracket — 3(n + 1) is 3n + 3, not 3n + 1.
(i) a₁₄ = a₁ + 13d = 3 + 13 × 7 = 3 + 91 = 94. (ii) d = 3 − 8 = −5. a₁₂ = 8 + 11 × (−5) = 8 − 55 = −47.
Write both terms with the formula: a₁₈ = a₁ + 17d = 367 and a₃₀ = a₁ + 29d = 499. Subtract: 12d = 499 − 367 = 132, so d = 11. Then a₁ = 367 − 17 × 11 = 367 − 187 = 180. The general term is an = 180 + (n − 1) × 11. We need an < 1000: 180 + 11(n − 1) < 1000 gives 11(n − 1) < 820, so n − 1 < 74.54…, i.e. n ≤ 75. Check: a₇₅ = 180 + 74 × 11 = 994 (< 1000) and a₇₆ = 1005. So 75 terms are less than 1000.
Here a₁ = 5 and d = 6, so an = 5 + (n − 1) × 6. Set it equal to 301: 5 + 6(n − 1) = 301 gives 6(n − 1) = 296, so n − 1 = 49.33…, which is not a whole number. A term number must be a natural number, so no, 301 is not a term (the sequence jumps from a₅₀ = 299 straight to a₅₁ = 305).
Three numbers are in A.P. exactly when twice the middle one equals the sum of the other two: 2(x + 8) = 2x + (3x + 1). The left side is 2x + 16 and the right side is 5x + 1, so 2x + 16 = 5x + 1 gives 15 = 3x, hence x = 5. Check: the numbers become 10, 13, 16 — an A.P. with d = 3.
Here a₁ = 3, d = 5, so an = 3 + (n − 1) × 5. Set it equal to 123: 5(n − 1) = 120, so n − 1 = 24 and n = 25. So 123 is the 25th term.
Here a₁ = 30, d = −0.5, so an = 30 + (n − 1)(−0.5). We need an < 0: 30 − 0.5(n − 1) < 0 gives 0.5(n − 1) > 30, so n − 1 > 60, i.e. n > 61. Check the boundary: a₆₁ = 30 − 30 = 0, which is not negative, and a₆₂ = 30 − 30.5 = −0.5. So the first negative term is the 62nd term.
a₇ = a₁ + 6d = 37 and a₂₁ = a₁ + 20d = 107. Subtract: 14d = 70, so d = 5. Then a₁ = 37 − 6 × 5 = 7. The A.P. is 7, 12, 17, … and a₁₀₀ = 7 + 99 × 5 = 7 + 495 = 502.
⚠ Book note (read this first): as printed, this statement can only hold for non-real values — with real distinct a, b, c the hypothesis is impossible, because the algebra below forces X² + Y² = 0 (see the last line). The proof below establishes the identity exactly as the book asks it, following the book's own algebra. Put X = a − c and Y = b − a, so that b − c = X + Y. The A.P. condition (twice the middle = sum of the neighbours) is 2/(b − c) = 1/(a − c) + 1/(b − a), i.e. 2/(X + Y) = 1/X + 1/Y = (X + Y)/(XY). Cross-multiplying gives 2XY = (X + Y)² = X² + 2XY + Y², so X² + Y² = 0, i.e. X² = −Y². Dividing both sides by XY gives X/Y = −Y/X. But X/Y = (a − c)/(b − a) and −Y/X = (a − b)/(a − c), so (a − b)/(a − c) = (a − c)/(b − a), as required. (Over the reals, X² + Y² = 0 would force X = Y = 0, contradicting the non-zero denominators — so the hypothesis has no real distinct solution; the identity itself is correct algebra.)
The three-digit multiples of 7 form an A.P.: 105, 112, 119, …, 994, with a₁ = 105 and d = 7. If there are n of them, the last term gives 994 = 105 + (n − 1) × 7, so (n − 1) × 7 = 889 and n − 1 = 127. Hence n = 128 three-digit numbers are divisible by 7.
First count the terms: 185 = 8 + (m − 1) × 3 gives (m − 1) × 3 = 177, so m − 1 = 59 and m = 60 terms in all. By the book's note, the nth term from the end is the (m − n + 1)th term from the start, so the 8th from the end is the (60 − 8 + 1) = 53rd term: a₅₃ = 8 + 52 × 3 = 8 + 156 = 164.
Look at the bases: 3, 10, 17, … form an A.P. with difference 7, so the nth base is 3 + (n − 1) × 7 = 7n − 4. Hence the nth term is ((7n − 4)/7)10. Is it an A.P.? For an A.P. we would need 2a₂ = a₁ + a₃, i.e. 2 × 1010 = 310 + 1710. But 2 × 1010 = 20,000,000,000 while 1710 alone is already about 2.0 × 10¹², far larger — the equality fails. So no, it is not an A.P. (the 10th powers grow far too fast for equal differences).
First A.P.: an = 3 + (n − 1) × 7 = 7n − 4. Second A.P.: bn = 63 + (n − 1) × 2 = 2n + 61. Setting them equal: 7n − 4 = 2n + 61 gives 5n = 65, so n = 13. Check: both 13th terms are 87.
Write ap = a₁ + (p − 1)d = q and aq = a₁ + (q − 1)d = p. Subtract: (p − q)d = q − p, so (p − q)d = −(p − q). Since p ≠ q (the terms are different positions), d = −1. Then a₁ = q − (p − 1)(−1) = q + p − 1. Hence an = a₁ + (n − 1)d = (p + q − 1) + (n − 1)(−1) = p + q − n, as required.
The A.P. condition gives 2/b = 1/a + 1/c = (c + a)/(ac). Taking reciprocals of both sides (flipping the fractions) gives b/2 = ac/(a + c), and doubling gives b = 2ac/(a + c). Note this is exactly the harmonic-mean formula — b here is the harmonic mean of a and c.
From Q17, the A.P. 1/a, 1/b, 1/c has middle term 1/b = (a + c)/(2ac). The common difference is the middle term minus the first: d = 1/b − 1/a = (a + c)/(2ac) − 1/a. Writing 1/a as 2c/(2ac) gives d = (a + c − 2c)/(2ac) = (a − c)/(2ac), as required.
Let d be the common difference. Then ak = a₁ + (k − 1)d and am = a₁ + (m − 1)d. Subtract: ak − am = (k − m)d, so d = (ak − am)/(k − m) (this needs k ≠ m, which is given since the terms are different). By the book's note, the nth term from the kth term is an = ak + (n − k)d. Substituting d gives an = ak + (n − k)((ak − am)/(k − m)), as required.
Rationalise one typical term: 1/(√ak + √ak+1) = (√ak+1 − √ak)/((ak+1 − ak)) = (√ak+1 − √ak)/d, since consecutive terms differ by d. Adding all n − 1 terms, the middle square roots cancel in pairs (telescoping): the sum is (√an − √a₁)/d. Now an − a₁ = (n − 1)d, and also an − a₁ = (√an − √a₁)(√an + √a₁), so (√an − √a₁)/d = (n − 1)/(√an + √a₁). Hence the sum equals (n − 1)/(√a₁ + √an), as required.
Equal roots means the discriminant is zero: (c − a)² − 4(b − c)(a − b) = 0. Expand: c² − 2ac + a² − 4(ab − b² − ac + bc) = a² + 4b² + c² + 2ac − 4ab − 4bc. This is exactly (a − 2b + c)² expanded (check: (a + c − 2b)² = a² + c² + 4b² + 2ac − 4ab − 4bc). So (a − 2b + c)² = 0, giving a − 2b + c = 0, i.e. 2b = a + c. That is precisely the condition for a, b, c to be in A.P.
Write the sides as a − d, a, a + d, with a + d the hypotenuse (the longest side). Pythagoras gives (a − d)² + a² = (a + d)². Expand: a² − 2ad + d² + a² = a² + 2ad + d², which simplifies to a² = 4ad, so a = 4d (d ≠ 0, else it is not a triangle). The sides are therefore 3d, 4d, 5d, in the ratio 3 : 4 : 5.
Let an = pn + q, where p and q are constants (a linear expression in n). Then the next term is an+1 = p(n + 1) + q = pn + q + p, so the difference an+1 − an = p, the same number for every n. A sequence whose consecutive differences are constant is an A.P. by definition, with common difference p. Hence the progression is an A.P.
(i) The A.M. is the average: [(2 + √3i) + (2 − √3i)]/2 = 4/2 = 2 (the imaginary parts cancel). (ii) [(a + b)² + (a − b)²]/2 = [(a² + 2ab + b²) + (a² − 2ab + b²)]/2 = (2a² + 2b²)/2 = a² + b².
Then a, 6, 11, 16, b is an A.P. The visible terms differ by 5, so d = 5. Stepping back once from 6 gives a = 6 − 5 = 1, and stepping forward once from 16 gives b = 16 + 5 = 21. Check: 1, 6, 11, 16, 21 is an A.P. with d = 5.
With five means inserted there are 7 terms in all, so d = (15/√2 − √2)/6. Now 15/√2 − √2 = (15 − 2)/√2 = 13/√2, so d = 13/(6√2) = 13√2/12. Adding d repeatedly to √2: first mean √2 + 13√2/12 = 25√2/12; second 19√2/6; third 17√2/4; fourth 16√2/3; fifth 77√2/12. Check: one more step gives 77√2/12 + 13√2/12 = 90√2/12 = 15√2/2 = 15/√2, as required.
Let the numbers be x and y. Then (x + y)/2 = 7, so x + y = 14, and xy = 45. So x and y are the roots of t² − 14t + 45 = 0, which factors as (t − 5)(t − 9) = 0. The numbers are 5 and 9.
With n means inserted, the full A.P. is a, A₁, …, An, b — that is n + 2 terms, with first term a and last term b. The last term formula gives b = a + (n + 2 − 1)d = a + (n + 1)d. Rearranging, (n + 1)d = b − a, so d = (b − a)/(n + 1), as required. The key point is the count: there are n + 1 steps from a to b, not n.
Here A = (a + b)/2. Then a − A = a − (a + b)/2 = (a − b)/2, and A − b = (a + b)/2 − b = (a − b)/2 as well — both gaps are half the total gap, as they must be for a mean. Squaring and adding: [(a − b)/2]² + [(a − b)/2]² = 2 × (a − b)²/4 = (a − b)²/2, as required.
The A.M. between a and b is (a + b)/2, so we need (an+1 + bn+1)/(an + bn) = (a + b)/2. Cross-multiply: 2(an+1 + bn+1) = (a + b)(an + bn) = an+1 + abn + ban + bn+1. Cancelling an+1 + bn+1 from both sides leaves an+1 + bn+1 = abn + ban, i.e. an(a − b) = bn(a − b). Since a ≠ b, divide by (a − b): an = bn, which for positive distinct a, b forces n = 0. Check: for n = 0 the expression is (a + b)/(1 + 1) = (a + b)/2, the A.M., as required.
(i) This is an A.P. with a₁ = 3, d = 3, so a₂₀ = 3 + 19 × 3 = 60. S₂₀ = (20/2)(3 + 60) = 10 × 63 = 630. (ii) The terms are 4/√5, 5/√5, 6/√5, …, so the kth term is (k + 3)/√5. Summing: Sn = (1/√5)[(1 + 2 + … + n) + 3n] = (1/√5)[n(n + 1)/2 + 3n] = n(n + 7)/(2√5).
(i) Last term known, so use Sn = (n/2)(a₁ + an): S₂₅ = (25/2)(4 + 100) = 25 × 52 = 1300. (ii) Use Sn = (n/2)[2a₁ + (n − 1)d]: S₂₀ = 10[80 + 19(−3)] = 10 × 23 = 230. (iii) First find a₁: an = a₁ + 20d gives 52 = a₁ + 20(−4), so a₁ = 132. Then S₂₁ = (21/2)(132 + 52) = 21 × 92 = 1932.
Sn = (n/2)[2a₁ + (n − 1)d] gives 1786 = (19/2)[2a₁ + 18 × 8] = (19/2)(2a₁ + 144). Divide by 19: 94 = (2a₁ + 144)/2, so 2a₁ + 144 = 188, giving 2a₁ = 44 and a₁ = 22.
Here a₁ = 96, d = −3, so Sn = (n/2)[192 + (n − 1)(−3)] = (n/2)(195 − 3n). Set equal to 1071: n(195 − 3n) = 2142, i.e. 3n² − 195n + 2142 = 0, i.e. n² − 65n + 714 = 0. The discriminant is 65² − 4 × 714 = 4225 − 2856 = 1369 = 37², so n = (65 ± 37)/2, giving n = 14 or n = 51. ⚠ Book note: both values genuinely work — S₁₄ = 1071, and after the terms turn negative the running sum climbs back to 1071 again at S₅₁. If the question means "the first time the sum reaches 1071", the answer is 14 terms.
Write the sides as a − d, a, a + d. The perimeter gives 3a = 36, so a = 12. Pythagoras, with a + d as hypotenuse: (12 − d)² + 12² = (12 + d)². Expand: 144 − 24d + d² + 144 = 144 + 24d + d², which simplifies to 144 = 48d, so d = 3. The sides are 9 cm, 12 cm, 15 cm.
(i) Group in threes: (3 + 5 − 7) = 1, (9 + 11 − 13) = 7, (15 + 17 − 19) = 13, … — the group sums form an A.P. with first term 1 and difference 6. The sum of n groups is (n/2)[2 × 1 + (n − 1) × 6] = (n/2)(6n − 4) = 3n² − 2n. (ii) Group in threes: (1 + 4 − 7) = −2, (10 + 13 − 16) = 7, (19 + 22 − 25) = 16, … — an A.P. with first term −2 and difference 9. The sum of n groups is (n/2)[2(−2) + (n − 1) × 9] = (n/2)(9n − 13) = (9n² − 13n)/2.
Sum the formula term by term: S₂₀ = Σr=120 (3r + 1) = 3 × Σr + Σ1 = 3 × (20 × 21/2) + 20 = 3 × 210 + 20 = 630 + 20 = 650.
a₅ = a₁ + 4d = 11 and a₉ = a₁ + 8d = 17. Subtract: 4d = 6, so d = 3/2. Then a₁ = 11 − 4 × (3/2) = 11 − 6 = 5. S₂₀ = (20/2)[2 × 5 + 19 × (3/2)] = 10[10 + 28.5] = 10 × 38.5 = 385.
Total of 1 to 1000: 1000 × 1001/2 = 500,500. Multiples of 2: 2 + 4 + … + 1000 = 2(1 + … + 500) = 2 × 125,250 = 250,500. Multiples of 5: 5(1 + … + 200) = 5 × 20,100 = 100,500. Multiples of both (i.e. of 10): 10(1 + … + 100) = 10 × 5,050 = 50,500. By inclusion–exclusion, multiples of 2 or 5 total 250,500 + 100,500 − 50,500 = 300,500. So the required sum is 500,500 − 300,500 = 200,000.
S₉ = (9/2)[2a₁ + 8d] = 9(a₁ + 4d) = 9a₅, so a₅ = 171/9 = 19. Also a₈ = a₁ + 7d = 31. Since a₈ − a₅ = 3d = 31 − 19 = 12, d = 4. Then a₁ = 19 − 4 × 4 = 3. The series is 3 + 7 + 11 + …
a₅ = a₁ + 4d = 21. S₆ = (6/2)[2a₁ + 5d] = 3(2a₁ + 5d) = 90, so 2a₁ + 5d = 30. Double the first equation: 2a₁ + 8d = 42. Subtract: 3d = 12, so d = 4, and a₁ = 21 − 16 = 5. Then a₁₈ = 5 + 17 × 4 = 5 + 68 = 73.
Write the numbers as a − d, a, a + d. Their sum is 3a = 24, so a = 8. Their product is (8 − d) × 8 × (8 + d) = 8(64 − d²) = 440, so 64 − d² = 55, giving d² = 9 and d = ±3. The numbers are 5, 8, 11 (in increasing order).
Here a₁ = 2, d = 4, so Sn = (n/2)[4 + (n − 1) × 4] = (n/2)(4n) = 2n². We need 2n² > 200, i.e. n² > 100, i.e. n > 10. The least such n is 11 (note S₁₀ = 200 exactly, which is not greater than 200, so 10 is not enough).
Write the numbers as a − 3d, a − d, a + d, a + 3d (the symmetric choice makes the sum simple). Sum: 4a = 32, so a = 8. Sum of squares: (8 − 3d)² + (8 − d)² + (8 + d)² + (8 + 3d)² = 4 × 64 + (9 + 1 + 1 + 9)d² = 256 + 20d² = 276, so 20d² = 20 and d² = 1, d = ±1. The numbers are 5, 7, 9, 11.
Write them as a − 2d, a − d, a, a + d, a + 2d. Sum: 5a = 25, so a = 5. Sum of squares: 5 × 25 + (4 + 1 + 0 + 1 + 4)d² = 125 + 10d² = 135, so d² = 1 and d = ±1. The numbers are 3, 4, 5, 6, 7.
The A.P. condition (twice the middle = sum of neighbours) gives 2/(c + a) = 1/(a + b) + 1/(b + c) = (a + 2b + c)/((a + b)(b + c)). Cross-multiply: 2(a + b)(b + c) = (c + a)(a + 2b + c). Expand the left: 2(ab + ac + b² + bc). Expand the right: (a + c)(a + c + 2b) = (a + c)² + 2b(a + c) = a² + 2ac + c² + 2ab + 2bc. Cancelling the common terms 2ab, 2ac, 2bc from both sides leaves 2b² = a² + c². That is exactly the condition for a², b², c² to be in A.P.
Let there be m terms and difference d. First four sum: (4/2)[2 × 11 + 3d] = 2(22 + 3d) = 56, so 22 + 3d = 28 and d = 2. The last four terms are am−3, …, am, with am−3 = 11 + (m − 4) × 2 = 2m + 3 and am = 11 + (m − 1) × 2 = 2m + 9. Their sum is (4/2)[(2m + 3) + (2m + 9)] = 2(4m + 12) = 8m + 24 = 112, so 8m = 88 and m = 11. Check: the A.P. is 11, 13, …, 31; first four sum to 56 and last four (25 + 27 + 29 + 31) to 112.
The common difference is d = b − a. If there are m terms, the last term gives c = a + (m − 1)d = a + (m − 1)(b − a), so (m − 1)(b − a) = c − a and m − 1 = (c − a)/(b − a). Hence m = (c − a + b − a)/(b − a) = (b + c − 2a)/(b − a). The sum is S = (m/2)(a + c) = (b + c − 2a)(c + a)/(2(b − a)), as required.
The n means together with a and b form an A.P. with n + 2 terms and difference d = (b − a)/(n + 1). The first mean is a + d and the last mean is b − d, so the sum of the n means is (n/2)[(a + d) + (b − d)] = (n/2)(a + b) = n × (a + b)/2. Since (a + b)/2 is the single A.M. between a and b, the sum of the n means is n times the single A.M.
The ratio is r = (−3)/(−6) = 1/2. Then a₆ = a₁r⁵ = (−6)(1/2)⁵ = −6/32 = −3/16.
The nth term is 3n (first term 3¹, second 3², …), so the 8th term is 3⁸ = 6561.
First sequence: an = 2n−1. Second: bn = 256 × (1/2)n−1 = 2⁸ × 2−(n−1) = 29−n. Setting them equal: 2n−1 = 29−n, so n − 1 = 9 − n, giving 2n = 10 and n = 5. Check: both 5th terms are 16.
(i) Multiply by 1/3 each time: 243, 81, 27, 9, 3. (ii) Multiply by −1/2 each time (signs alternate): 579, −579/2, 579/4, −579/8, 579/16.
First find r: (2i)/(1 + i) = 2i(1 − i)/((1 + i)(1 − i)) = (2i + 2)/2 = 1 + i. Check with the next pair: (−2 + 2i)/(2i) = (−2 + 2i)(−i)/2 = (2i + 2)/2 = 1 + i. Good, r = 1 + i. Then a₁₂ = (1 + i)(1 + i)¹¹ = (1 + i)¹². Since (1 + i)² = 2i, (1 + i)¹² = (2i)⁶ = 64 × i⁶ = 64 × (−1) = −64.
a₄ = a₁r³ = 54 and a₉ = a₁r⁸ = 13122. Divide: r⁵ = 13122/54 = 243, so r = 3 (243 = 3⁵). Then a₁ = 54/r³ = 54/27 = 2. The G.P. is 2, 6, 18, … and its general term is an = 2 × 3n−1.
Write the G.P. as a, ar, ar², ar³ with ratio r. (i) a − b = a(1 − r), b − c = ar(1 − r), c − d = ar²(1 − r): each is r times the previous, so they are in G.P. with ratio r (for r ≠ 1; if r = 1 all three differences are 0). (ii) a² − b² = a²(1 − r²), b² − c² = a²r²(1 − r²), c² − d² = a²r⁴(1 − r²): each is r² times the previous, so they are in G.P. with ratio r². (iii) a² + b² = a²(1 + r²), b² + c² = a²r²(1 + r²), c² + d² = a²r⁴(1 + r²): again each is r² times the previous, so they are in G.P. with ratio r².
Let the first term be a and ratio r. Then ap+q = arp+q−1 = m and ap−q = arp−q−1 = n. Multiply the two equations: mn = a²r2p−2 = (arp−1)². But arp−1 is exactly the pth term, so the pth term is √(mn) (the positive root, for a G.P. of positive terms; in general it is ±√(mn), the sign depending on the terms).
Write the numbers as a/r, a, ar. Their product is a³ = 216, so a = 6. Their sum is 6(1/r + 1 + r) = 26, so 1/r + r = 26/6 − 1 = 10/3. Multiply by r: 1 + r² = (10/3)r, i.e. 3r² − 10r + 3 = 0, i.e. (3r − 1)(r − 3) = 0, so r = 3 or r = 1/3. Either way the numbers are 2, 6, 18.
Let the first term be a and ratio r. Then a₂ = ar = 9 and a₃ = ar² = a². From ar² = a² (and a ≠ 0), r² = a. Then ar = 9 becomes r² × r = 9, i.e. r³ = 9, so r = ∛9 and a = r². The 6th term is a₆ = ar⁵ = (ar) × r⁴ = 9 × (∛9)⁴ = 9 × 9∛9 = 81∛9 (about 168.5). Check: a₃ = a² = (∛9)⁴ = 9∛9, and a₂ × r = 9 × ∛9 = 9∛9. ✓
The common ratio is (1/b)/(1/a) = a/b, and also (1/c)/(1/b) = b/c. Since the three reciprocals are in G.P., the middle term squared equals the product of its neighbours: (1/b)² = (1/a)(1/c), so b² = ac and b = ±√(ac). Hence the ratio a/b = a/(±√(ac)) = ±√(a/c), as required (the ± appears because b may be either square root of ac).
Write the A.P. terms as a − d, a, a + d. Their sum is 3a = 21, so a = 7. After subtraction the numbers are (7 − d − 1), (7 − 4), (7 + d − 3) = (6 − d), 3, (4 + d), in G.P. So 3² = (6 − d)(4 + d), i.e. 9 = 24 + 2d − d², i.e. d² − 2d − 15 = 0, i.e. (d − 5)(d + 3) = 0. So d = 5 or d = −3. The original numbers are 2, 7, 12 (for d = 5) or 10, 7, 4 (for d = −3); both work — the resulting G.P.s are 1, 3, 9 and 9, 3, 1 respectively.
Write the A.P. numbers as a − d, a, a + d. Their sum is 3a = 6, so a = 2. After the increases: (3 − d), 6, (17 + d), in G.P. So 6² = (3 − d)(17 + d), i.e. 36 = 51 − 14d − d², i.e. d² + 14d − 15 = 0, i.e. (d + 15)(d − 1) = 0. So d = 1 or d = −15. The original numbers are 1, 2, 3 (for d = 1) or 17, 2, −13 (for d = −15); the resulting G.P.s are 2, 6, 18 and 18, 6, 2 respectively.
Let the first term be a and ratio r. Then arp−1 = q and arq−1 = p. The cleanest proof uses logarithms. Let A = ln a, R = ln r. Then A + (p − 1)R = ln q and A + (q − 1)R = ln p. Subtracting gives (p − q)R = ln q − ln p. The log of the (p + q)th term is A + (p + q − 1)R = [A + (p − 1)R] + qR = ln q + q(ln q − ln p)/(p − q) = [p ln q − q ln p]/(p − q) = ln[(qp/pq)1/(p−q)]. Hence the (p + q)th term is (qp/pq)1/(p−q), as printed in the book (the book's ÷ sign between qp and pq is easy to misread as +).
The G.P. condition on the first three terms is (2a + 2)² = a(3a + 3), i.e. 4(a + 1)² = 3a(a + 1), i.e. (a + 1)(4a + 4 − 3a) = (a + 1)(a + 4) = 0. So a = −1 or a = −4. The case a = −1 gives −1, 0, 0, … which is not a genuine G.P. (the ratio is undefined), so discard it. For a = −4 the terms are −4, −6, −9, … with ratio r = (−6)/(−4) = 3/2. The fifth term is a₅ = a₁r⁴ = (−4)(3/2)⁴ = (−4)(81/16) = −81/4. (Note: the book asks for the G.P. continuation here, not the "na + n" pattern of the printed terms — the G.P. terms after the third are −27/2, −81/4, ….)
(i) G = ±√((−2)(8)) = ±√(−16) = ±4i. (ii) G = ±√((−2i)(8i)) = ±√(−16i²) = ±√16 = ±4. (iii) G = ±√(6 × 9) = ±√54 = ±3√6.
With four means inserted there are 6 terms in all, so 96 = 3r⁵, giving r⁵ = 32 and, for real means, r = 2. The means are 3 × 2 = 6, 6 × 2 = 12, 12 × 2 = 24, 24 × 2 = 48. Check: one more step gives 96. (The word "real" matters: r⁵ = 32 also has complex roots, which would give complex means — the book asks you to leave those out.)
Since x ≠ y are positive, √x ≠ √y, so (√x − √y)² > 0. Expand: x − 2√(xy) + y > 0, so x + y > 2√(xy). Divide by 2: (x + y)/2 > √(xy). The left side is the A.M. and the right side is the (positive) G.M., so G.M. < A.M., as required.
The positive G.M. is √(ab), so we need (an + bn)/(an−1 + bn−1) = √(ab). Cross-multiply: an + bn = √(ab)(an−1 + bn−1) = an−1/2b1/2 + a1/2bn−1/2. Bring everything to one side and factor: an−1/2(a1/2 − b1/2) − bn−1/2(a1/2 − b1/2) = (a1/2 − b1/2)(an−1/2 − bn−1/2) = 0. Since a and b are distinct positives, a1/2 ≠ b1/2, so we need an−1/2 = bn−1/2, which forces n − 1/2 = 0, i.e. n = 1/2. Check: for n = 1/2 the expression is (√a + √b)/(1/√a + 1/√b) = √(ab). ✓
Let the numbers be x and y. Then (x + y)/2 − √(xy) = 2 and x + y = 20, so 10 − √(xy) = 2, giving √(xy) = 8 and xy = 64. So x and y are roots of t² − 20t + 64 = 0, i.e. (t − 4)(t − 16) = 0. The numbers are 4 and 16 (both positive integers, as stated).
Then x + y = 10 and √(xy) = 4, so xy = 16. The numbers are roots of t² − 10t + 16 = 0, i.e. (t − 2)(t − 8) = 0. The numbers are 2 and 8.
The condition is (a + b)/2 = 2√(ab), i.e. a + b = 4√(ab). Divide by b and put t = √(a/b) > 0: then a/b = t² and the equation becomes t² + 1 = 4t, i.e. t² − 4t + 1 = 0, so t = 2 ± √3. Hence a/b = t² = (2 + √3)² = 7 + 4√3 (taking a > b). Now (2 + √3)/(2 − √3) = (2 + √3)²/((2)² − (√3)²) = (7 + 4√3)/1 = 7 + 4√3. So a/b = (2 + √3)/(2 − √3), i.e. a : b = 2 + √3 : 2 − √3, as required.
Let the two numbers be a and b. The two A.M.s give the A.P. a, p, q, b, with difference d = (b − a)/3, so p = a + d and q = a + 2d. Then 2p − q = 2(a + d) − (a + 2d) = a, and 2q − p = 2(a + 2d) − (a + d) = a + 3d = b. Hence (2p − q)(2q − p) = ab. But G is the G.M. between a and b, so G² = ab. Therefore G² = (2p − q)(2q − p), as required.
Here a₁ = 1, r = 1/3. S₁₅ = (1 − (1/3)¹⁵)/(1 − 1/3) = (3/2)(1 − 3−15) = (3¹⁵ − 1)/(2 × 3¹⁴) = 7174453/4782969, about 1.4999999 — just short of the sum to infinity, 3/2.
a₃ = a₁r² = 16 and a₆ = a₁r⁵ = −128. Divide: r³ = −128/16 = −8, so r = −2. Then a₁ = 16/r² = 16/4 = 4. S₇ = 4(1 − (−2)⁷)/(1 − (−2)) = 4(1 + 128)/3 = 4 × 129/3 = 172.
(i) The kth term, 0.222… (k twos), equals (2/9)(1 − 10−k). Summing: Sn = (2/9)[n − Σ10−k] = (2/9)[n − (1/10)(1 − 10−n)/(1 − 1/10)] = (2/9)[n − (1 − 10−n)/9] = (2n/9) − (2/81)(1 − 10−n). (ii) The kth term is (10k − 1)/3. Summing: Sn = (1/3)[(10 + 10² + … + 10n) − n] = (1/3)[10(10n − 1)/9 − n] = (10n+1 − 10 − 9n)/27.
(i) The term with index k (starting k = 0) is 1 + a + … — more precisely the (k + 1)th term is (ak+1 − bk+1)/(a − b), the sum of a G.P. So Sn = (1/(a − b))[Σk=0n−1 ak+1 − Σk=0n−1 bk+1] = (1/(a − b))[a(1 − an)/(1 − a) − b(1 − bn)/(1 − b)] (for a, b ≠ 1). (ii) The mth term is rm(1 + k + … + km−1) = rm(km − 1)/(k − 1) = [(rk)m − rm]/(k − 1). Summing two G.P.s: Sn = (1/(k − 1))[rk(1 − (rk)n)/(1 − rk) − r(1 − rn)/(1 − r)] (for k ≠ 1, r ≠ 1, rk ≠ 1).
First find r: (1 − i)/2, and check: (1/i)/(1 − i) = (−i)/(1 − i) = (−i)(1 + i)/2 = (1 − i)/2. Good, r = (1 − i)/2. Now r² = (1 − 2i + i²)/4 = −i/2, r⁴ = (−i/2)² = −1/4, r⁸ = 1/16. Then S₈ = 2(1 − r⁸)/(1 − r) = 2(15/16)/((1 + i)/2) = (15/8) × (2/(1 + i)) = (15/4) × (1 − i)/2 = 15(1 − i)/8.
Sum of the first n terms: S = a₁(1 − rn)/(1 − r). The terms from the (n + 1)th to the (2n)th form a G.P. with first term a₁rn and n terms, so their sum is a₁rn(1 − rn)/(1 − r). The ratio of the two sums is [a₁(1 − rn)/(1 − r)] ÷ [a₁rn(1 − rn)/(1 − r)] = 1/rn, as required.
The A.P. part has 8th term 1 + 7 × 3 = 22. The G.P. part has 8th term 5 × 2⁷ = 5 × 128 = 640. The A.G.P. term is the product: 22 × 640 = 14080.
The A.P. part has nth term 3 + (n − 1) × 4 = 4n − 1. The G.P. part has nth term 2 × 3n−1. Multiply: the nth term of the A.G.P. is 2(4n − 1)3n−1.
⚠ Index note (derived, not printed): the book gives the parts in shifted index form, so first unshift them. For the A.P. part, an+1 = 2n + 5 means a₁ (n = 0) = 5, a₂ = 7, a₃ = 9, …, i.e. an = 2n + 3. For the G.P. part, putting k = n − 2 gives bk = (1/9)(−3)k+2 = (−3)k, so bn = (−3)n. Hence the nth term of the A.G.P. is (2n + 3)(−3)n. First three terms: n = 1: 5 × (−3) = −15; n = 2: 7 × 9 = 63; n = 3: 9 × (−27) = −243. Sum: −15 + 63 − 243 = −195.
(i) The kth term is (2k − 1)2k. Write Sn = Σ(2k − 1)2k and 2Sn = Σ(2k − 1)2k+1 = Σk=2n+1(2k − 3)2k. Subtract: −Sn = 2 + Σk=2n 2·2k − (2n − 1)2n+1 = 2 + 2(2n+1 − 4) − (2n − 1)2n+1 = (3 − 2n)2n+1 − 6. Hence Sn = (2n − 3)2n+1 + 6 (check n = 1: −4 + 6 = 2 ✓). (ii) The kth term is 2k·3k. With the same trick, 2Sn = 3 + (2n − 1)3n+1 (the middle terms telescope: each step contributes −2·3k), so Sn = [3 + (2n − 1)3n+1]/2 (check n = 2: (3 + 3 × 27)/2 = 42 = 6 + 36 ✓). (iii) A.P. part 2, 5, 8, … (first term 2, d = 3), G.P. part 1, 1/4, … (r = 1/4). By the book's A.G.P. formula, Sn = 2/(3/4) + 3(1/4)/(3/4)² − 3(1/4)n/(3/4)² − (3n − 1)(1/4)n/(3/4) = 8/3 + 4/3 − (1/4)n[(16/3) + (4/3)(3n − 1)] = 4 − (n + 1)(1/4)n−1 (check n = 2: 4 − 3/4 = 13/4 = 2 + 5/4 ✓). (iv) A.P. part 1, 3, 5, … (d = 2), r = 1/5. Sn = 1/(4/5) + 2(1/5)/(4/5)² − 2(1/5)n/(4/5)² − (2n − 1)(1/5)n/(4/5) = 15/8 − (5/8)(4n + 3)(1/5)n (check n = 2: 15/8 − 55/200 = 8/5 = 1 + 3/5 ✓). (v) A.P. part 1, 4, 7, … (d = 3), r = 1/3. Sn = 1/(2/3) + 3(1/3)/(2/3)² − 3(1/3)n/(2/3)² − (3n − 2)(1/3)n/(2/3) = 15/4 − (3/4)(6n + 5)(1/3)n (check n = 2: 15/4 − 17/12 = 7/3 = 1 + 4/3 ✓).
(i) A.P. part 1, 3, 5, … (a = 1, d = 2), G.P. part 1, 1/2, … (b = 1, r = 1/2). S∞ = ab/(1 − r) + dbr/(1 − r)² = 1/(1/2) + 2(1/2)/(1/2)² = 2 + 4 = 6. (ii) A.P. part 2, 5, 8, … (a = 2, d = 3), r = 1/3. S∞ = 2/(2/3) + 3(1/3)/(2/3)² = 3 + 9/4 = 21/4.
Write every factor as a power of 2: 21/2 = 21/2, 41/4 = 22/4 = 21/2, 81/8 = 23/8, 161/16 = 24/16 = 21/4, …. In general the kth factor is 2k/2k. Multiplying adds the exponents: total exponent = Σk=1∞ k/2k. This A.G.P. sum is (1/2)/(1 − 1/2)² = 2 (by the S∞ formula with a = d = 1, r = 1/2, or by the multiply-and-subtract trick). Hence the product is 2² = 4.
Write every factor as a power of 2: √4 = 2, ⁴√16 = 2, ⁸√64 = 26/8 = 23/4, ¹⁶√256 = 28/16 = 21/2, …. In general the kth factor is (4k)1/2k = 2k/2k−1. Adding exponents: Σk=1∞ k/2k−1 = 2 × Σ k/2k = 2 × 2 = 4. Hence the product is 2⁴ = 16.
The kth term is 2k·xk−1, so Sn = 2Σk=1n kxk−1. Let T = Σkxk−1. Then xT = Σkxk. Subtracting, (1 − x)T = (1 + x + … + xn−1) − nxn = (1 − xn)/(1 − x) − nxn, so T = [1 − (n + 1)xn + nxn+1]/(1 − x)². Hence Sn = 2[1 − (n + 1)xn + nxn+1]/(1 − x)².
Put r = (2n + 1)/(2n − 1) (here n is fixed for the whole sum; the terms are 1·r, 3·r², 5·r³, …, i.e. the kth term is (2k − 1)rk). Let S = Σk=1n(2k − 1)rk. Then rS = Σ(2k − 1)rk+1, and subtracting gives (1 − r)S = r + Σk=2n2rk − (2n − 1)rn+1 = r + 2r²(1 − rn−1)/(1 − r) − (2n − 1)rn+1. Now 1 − r = −2/(2n − 1), so 2/(1 − r) = −(2n − 1), and the middle piece becomes −(2n − 1)r²(1 − rn−1) = −(2n − 1)r² + (2n − 1)rn+1. The rn+1 terms cancel, leaving (1 − r)S = r − (2n − 1)r². Since r = (2n + 1)/(2n − 1), we have r − (2n − 1)r² = [(2n + 1) − (2n + 1)²]/(2n − 1) = (2n + 1)(−2n)/(2n − 1). Dividing by (1 − r) = −2/(2n − 1) gives S = n(2n + 1). Check n = 2: r = 5/3, sum = 5/3 + 3 × 25/9 = 10 = 2 × 5 ✓.
Put r = 1 + 1/n = (n + 1)/n. The sum is S = Σk=1n k·rk−1. Then rS = Σk=1n k·rk, and subtracting: (1 − r)S = (1 + r + … + rn−1) − nrn = (1 − rn)/(1 − r) − nrn. Since 1 − r = −1/n, we have (1 − rn)/(1 − r) = −n(1 − rn) = n(rn − 1). So (1 − r)S = n(rn − 1) − nrn = −n, and S = (−n)/(1 − r) = (−n)/(−1/n) = n², as required.
This is an A.G.P. with a = 2, d = 3, b = 1, r = x. By the book's formula, Sn = 2/(1 − x) + 3x/(1 − x)² − 3xn/(1 − x)² − (3n − 1)xn/(1 − x). Put over (1 − x)²: the first two terms give (2 + x)/(1 − x)², and the last two give −xn[3 + (3n − 1)(1 − x)]/(1 − x)² = −xn[3n + 2 − (3n − 1)x]/(1 − x)². So Sn = [(2 + x) − xn(3n + 2 − (3n − 1)x)]/(1 − x)². If |x| < 1, then xn → 0 and nxn → 0 as n → ∞, so S∞ = (2 + x)/(1 − x)².
(i) Reciprocals: 3, 5, 7, … — an A.P. with d = 2. Its 9th term is 3 + 8 × 2 = 19, so the 9th term of the H.P. is 1/19. (ii) Reciprocals: −5, −3, −1, … — an A.P. with d = 2. Its 9th term is −5 + 8 × 2 = 11, so the 9th term of the H.P. is 1/11.
(i) Work with reciprocals: −5/2 and 13/2. Insert five A.M.s between them: with 7 terms in all, d = (13/2 − (−5/2))/6 = 9/6 = 3/2. The A.M.s are −1, 1/2, 2, 7/2, 5. Taking reciprocals gives the harmonic means: −1, 2, 1/2, 2/7, 1/5. (ii) Reciprocals: 4 and 24. Five A.M.s between them: d = (24 − 4)/6 = 10/3. The A.M.s are 22/3, 32/3, 14, 52/3, 62/3. Taking reciprocals: 3/22, 3/32, 1/14, 3/52, 3/62.
Reciprocals form an A.P. with first term −3 and fifth term 5. Then −3 + 4d = 5, so d = 2. The 9th term of the reciprocal A.P. is −3 + 8 × 2 = 13, so the 9th term of the H.P. is 1/13.
The H.M. formula gives 5 = 2 × 2 × b/(2 + b) = 4b/(2 + b). Cross-multiply: 5(2 + b) = 4b, i.e. 10 + 5b = 4b, so b = −10. (Equivalently, via reciprocals: 1/5 = (1/2 + 1/b)/2 gives 1/b = −1/10.) The negative answer is fine — nothing in the question required b positive.
Their reciprocals k, 2k + 1, 4k − 1 must be in A.P., so twice the middle equals the sum of the neighbours: 2(2k + 1) = k + (4k − 1), i.e. 4k + 2 = 5k − 1, giving k = 3.
The H.M. between a and b is 2ab/(a + b), so we need (an+1 + bn+1)/(an + bn) = 2ab/(a + b). Cross-multiply: (a + b)(an+1 + bn+1) = 2ab(an + bn). Expand the left: an+2 + abn+1 + ban+1 + bn+2. Subtracting the right side gives an+2 − an+1b − abn+1 + bn+2 = (a − b)(an+1 − bn+1) = 0. For distinct a, b this forces an+1 = bn+1, i.e. n + 1 = 0, so n = −1. Check: for n = −1 the expression is (1 + 1)/(1/a + 1/b) = 2ab/(a + b). ✓
It is enough to show the reciprocals 1/(a + b), 1/(c + a), 1/(b + c) are in A.P., i.e. 2/(c + a) = 1/(a + b) + 1/(b + c) = (a + 2b + c)/((a + b)(b + c)). Cross-multiply: 2(a + b)(b + c) = (a + c)(a + 2b + c). The left expands to 2ab + 2ac + 2b² + 2bc; the right expands to (a + c)² + 2b(a + c) = a² + 2ac + c² + 2ab + 2bc. Cancelling common terms leaves 2b² = a² + c², which is exactly the given fact that a², b², c² are in A.P. So the reciprocals are in A.P., and therefore a + b, c + a, b + c are in H.P.
Let the numbers be x and y. Then (x + y)/2 = 9/2, so x + y = 9, and 2xy/(x + y) = 4, so xy = 2 × 9 = 18. The numbers are roots of t² − 9t + 18 = 0, i.e. (t − 3)(t − 6) = 0. The numbers are 3 and 6.
G.M. = 4 gives xy = 16. H.M. = 16/5 gives 2xy/(x + y) = 16/5, so 32/(x + y) = 16/5 and x + y = 10. The numbers are roots of t² − 10t + 16 = 0, i.e. (t − 2)(t − 8) = 0. The numbers are 2 and 8.
The A.P. condition is 2(c + a − b)/b = (b + c − a)/a + (a + b − c)/c. Multiply through by abc: the left side becomes 2ac(c + a), and the right side becomes bc(b + c − a) + ab(a + b − c). So 2ac(c + a) = bc(b + c − a) + ab(a + b − c) = b²c + bc² − abc + a²b + ab² − abc. Bring everything to one side: 2ac² + 2a²c − b²c − bc² − a²b − ab² + 2abc = 0. This factors (check by expanding) as (2ac − b(a + c))(a + b + c) = 0: indeed (2ac − ab − bc)(a + b + c) = 2a²c + 2abc + 2ac² − a²b − ab² − abc − abc − b²c − bc² = 2a²c + 2ac² − a²b − ab² − b²c − bc², and adding the remaining 2abc gives exactly our expression. Since a + b + c ≠ 0, we get 2ac = b(a + c), i.e. 2/b = 1/a + 1/c. So 1/a, 1/b, 1/c are in A.P., which means a, b, c are in H.P.
Since a, b, c, d are in H.P., their reciprocals are in A.P. with some common difference D: 1/b = 1/a + D, 1/c = 1/a + 2D, 1/d = 1/a + 3D. Now a − b = ab(1/b − 1/a)·(−1)… compute directly: 1/b − 1/a = (a − b)/(ab) = D, so a − b = abD. Similarly b − c = bcD, c − d = cdD, and a − d: 1/d − 1/a = (a − d)/(ad) = 3D, so a − d = 3adD. Hence 3(a − b)(c − d) = 3(abD)(cdD) = 3abcdD², and (b − c)(a − d) = (bcD)(3adD) = 3abcdD². The two sides are equal, as required.
Let the two numbers be a and b. The A.M.s form the A.P. a, A₁, A₂, b, so A₁ + A₂ = (a + d) + (a + 2d) = 2a + 3d = a + b (since b = a + 3d). The G.M.s form the G.P. a, G₁, G₂, b, so G₁G₂ = (ar)(ar²) = a²r³ = ab (since b = ar³). Hence the left side is (a + b)/(ab) = 1/a + 1/b. For the H.M.s, the reciprocals 1/H₁, 1/H₂ are the two A.M.s between 1/a and 1/b, so by the same addition, 1/H₁ + 1/H₂ = 1/a + 1/b. But (H₁ + H₂)/(H₁H₂) = 1/H₂ + 1/H₁, so the right side also equals 1/a + 1/b. The two sides are equal, as required.
Let the numbers be x and y, with sum s = x + y. Then A = s/2, G² = xy, and H.M. = 2xy/s = 4, so xy = 2s. The relation 2A + G² = 27 becomes s + 2s = 27, so s = 9 and xy = 18. The numbers are roots of t² − 9t + 18 = 0, i.e. (t − 3)(t − 6) = 0. The numbers are 3 and 6.
From a, b, c in A.P.: a = 2b − c. From b, c, d in H.P.: 1/b, 1/c, 1/d are in A.P., so 2/c = 1/b + 1/d = (b + d)/(bd), i.e. 2bd = c(b + d). Now ad = (2b − c)d = 2bd − cd = c(b + d) − cd = bc. Hence ad = bc, as required.
It is enough to show their reciprocals are in A.P. Now 1/(loga x) = logx a, and similarly for b and c. Since a, b, c are in G.P., b² = ac, so taking logs to base x: 2 logx b = logx a + logx c. Hence logx a, logx b, logx c are in A.P., so their reciprocals — loga x, logb x, logc x — are in H.P. (This assumes a, b, c > 0 and ≠ 1, and x > 0, x ≠ 1, so the logs are defined.)
(i) Since 1/a, 1/b, 1/c are in A.P. with common difference, say, D: 1/b − 1/a = (a − b)/(ab) = D and 1/c − 1/b = (b − c)/(bc) = D. Equate: (a − b)/(ab) = (b − c)/(bc), and cross-multiplying by abc gives c(a − b) = a(b − c), i.e. (a − b)/(b − c) = a/c. (ii) From (i), c(a − b) = a(b − c), i.e. ac − bc = ab − ac, so 2ac = b(a + c) and b = 2ac/(a + c). Then a − 2b + c = (a + c) − 4ac/(a + c) = ((a + c)² − 4ac)/(a + c) = (a − c)²/(a + c). Multiply by (a + c): (a − c)² = (a + c)(a − 2b + c), as required.
Their reciprocals must be in A.P.: 2/(5 + x) = 1/(2 + x) + 1/(9 + x) = (11 + 2x)/((2 + x)(9 + x)). Cross-multiply: 2(2 + x)(9 + x) = (5 + x)(11 + 2x). Expand: 2(x² + 11x + 18) = 2x² + 21x + 55, i.e. 2x² + 22x + 36 = 2x² + 21x + 55, giving x = 19. Check: the numbers are 21, 24, 28, with reciprocals 1/21, 1/24, 1/28 — differences −1/168 each. ✓
Equal roots means discriminant zero: b²(c − a)² − 4ac(b − c)(a − b) = 0. Put s = a + c and p = ac. Then (a − b)(b − c) = b(a + c) − b² − ac = bs − b² − p, and (c − a)² = s² − 4p. The discriminant equation becomes b²(s² − 4p) − 4p(bs − b² − p) = 0, i.e. b²s² − 4b²p − 4pbs + 4pb² + 4p² = 0, i.e. b²s² − 4pbs + 4p² = 0, i.e. (bs − 2p)² = 0. So bs = 2p, i.e. b(a + c) = 2ac, i.e. 2/b = 1/a + 1/c. Hence 1/a, 1/b, 1/c are in A.P., so a, b, c are in H.P.
The method is the same throughout: write the kth term as a polynomial in k, then sum with Σk = n(n + 1)/2, Σk² = n(n + 1)(2n + 1)/6, Σk³ = [n(n + 1)/2]². (i) Term k(2k + 1) = 2k² + k: S = 2Σk² + Σk = n(n + 1)(2n + 1)/3 + n(n + 1)/2 = n(n + 1)(4n + 5)/6. (ii) Term k(3k + 2) = 3k² + 2k: S = 3Σk² + 2Σk = n(n + 1)(2n + 3)/2. (iii) Term k(3k − 1) = 3k² − k: S = 3Σk² − Σk = n²(n + 1). (iv) Term k(k + 2)(k + 4) = k³ + 6k² + 8k: S = Σk³ + 6Σk² + 8Σk = n(n + 1)(n + 4)(n + 5)/4. (v) Term k(k + 1)(3k + 1) = 3k³ + 4k² + k: S = 3Σk³ + 4Σk² + Σk = n(n + 1)(9n + 7)(n + 2)/12. (vi) Term (2k)² = 4k²: S = 4Σk² = 2n(n + 1)(2n + 1)/3. (vii) Term (3k)² = 9k²: S = 9Σk² = 3n(n + 1)(2n + 1)/2. (viii) Term (3k + 1)k² = 3k³ + k²: S = 3Σk³ + Σk² = n(n + 1)(9n² + 13n + 2)/12. (ix) The kth bracket is the sum of k terms of the A.P. 3, 7, 11, …, i.e. (k/2)[6 + 4(k − 1)] = 2k² + k — the same as (i) — so S = n(n + 1)(4n + 5)/6. (x) The kth bracket is Σi=1k i² = k(k + 1)(2k + 1)/6 = (2k³ + 3k² + k)/6: S = (1/6)[2Σk³ + 3Σk² + Σk] = n(n + 1)²(n + 2)/12.
(i) Pair consecutive terms: (2k − 1)² − (2k)² = (4k² − 4k + 1) − 4k² = −(4k − 1). Summing k = 1 to n: S = −Σ(4k − 1) = −[4 × n(n + 1)/2 − n] = −[2n(n + 1) − n] = −n(2n + 1). (ii) The kth term is [k(k + 1)(2k + 1)/6]/k = (k + 1)(2k + 1)/6 = (2k² + 3k + 1)/6. Summing: S = (1/6)[2Σk² + 3Σk + n] = (1/6)[n(n + 1)(2n + 1)/3 + 3n(n + 1)/2 + n] = n(4n² + 15n + 17)/36 (check n = 2: 2 × (16 + 30 + 17)/36 = 126/36 = 7/2 = 1 + 5/2 ✓).
(i) Sn = 5Σk² + 2Σk + 3n = 5n(n + 1)(2n + 1)/6 + n(n + 1) + 3n = n(10n² + 21n + 29)/6. (ii) Sn = Σk² + 2Σk − 3n = n(n + 1)(2n + 1)/6 + n(n + 1) − 3n = n(2n² + 9n − 11)/6.
(i) Replace n by 2n in the summation: S2n = 3Σk=12nk² + 5Σk=12nk + 2(2n) = 3 × (2n)(2n + 1)(4n + 1)/6 + 5 × (2n)(2n + 1)/2 + 4n = n(2n + 1)(4n + 1) + 5n(2n + 1) + 4n = 2n(4n² + 8n + 5). (ii) S2n = Σk=12n(k² + k − 2) = (2n)(2n + 1)(4n + 1)/6 + (2n)(2n + 1)/2 − 4n = (4n/3)(2n − 1)(n + 2).
The instalments form an A.P. with n = 40, d = 10 and S₄₀ = 10400. S₄₀ = (40/2)[2a₁ + 39 × 10] = 20(2a₁ + 390) = 10400, so 2a₁ + 390 = 520, giving 2a₁ = 130 and a₁ = Rs. 65.
Compound growth is a G.P. with ratio 1.06: amount = 150000 × (1.06)⁸. Now (1.06)⁸ = 1.593848…, so the amount is 150000 × 1.593848 = Rs. 239,077 (about, to the nearest rupee).
Geometric growth: 4084101 = 3200000 × (1 + r)⁵, so (1 + r)⁵ = 4084101/3200000 = 1.276281… Now 1.05⁵ = 1.2762815625, so 1 + r = 1.05 and the rate of increase is 5% per year.
⚠ Assumption stated: the book does not say whether each yearly payment is made at the start or the end of the year, so this answer uses the standard ordinary-annuity reading — Rs. 5000 paid at the end of each year, each then earning compound interest for the remaining years. The total is 5000 × [(1.05²⁰ − 1)/0.05] (the sum of a G.P. with first term 5000 × 1.05⁰… — precisely, the payments grow as 5000(1.05)¹⁹ + … + 5000, whose sum is 5000[(1.05²⁰ − 1)/0.05]). Since 1.05²⁰ = 2.653298, the bracket is (2.653298 − 1)/0.05 = 33.06596, giving Rs. 165,330 (about). (If instead each payment were made at the start of the year, every payment would earn one extra year and the total would be 165,330 × 1.05 ≈ Rs. 173,596.)
The amount falls by 5 gallons a week — an A.P. with difference −5: after w weeks the tank holds 2000 − 5w gallons. (i) After 20 weeks: 2000 − 5 × 20 = 1900 gallons. (ii) Half-full means 1000 gallons: 2000 − 5w = 1000 gives 5w = 1000, so w = 200 weeks. (iii) Empty means 0 gallons: 2000 − 5w = 0 gives w = 400 weeks.
(i) In 12 months the extra production is 12 × 1.4 = 16.8 million doses, so the total after a year is 7 + 16.8 = 23.8 million doses. (ii) Here n counts the number of months that have passed since "to date". The total after n months is an = 7 + 1.4n (million doses) — an A.P. with first value 7 and difference 1.4. (iii) a₁₀ = 7 + 1.4 × 10 = 21 million doses: that is the total number of doses produced (7 million already made plus 14 million in ten months) ten months from now.
(i) Each minute's count is 0.8 times the previous one — a G.P. with a₁ = 100, r = 0.8: 100, 80, 64, 51.2, …, with nth term 100 × (0.8)n−1. (ii) S₁₅ = 100(1 − 0.8¹⁵)/(1 − 0.8) = 500 × (1 − 0.03518) = 500 × 0.96482 = 482.41…, so about 482 vehicles in 15 minutes. (iii) As t → ∞, the sum tends to S∞ = 100/(1 − 0.8) = 500 vehicles — no matter how long you wait, the total can never pass 500.
Simple interest for one year is 5000 × 8/100 = Rs. 400, and it is the same every year, so the interest amounts at the ends of years 1, 2, 3, … are 400, 800, 1200, … — the interest accumulated after each year (as in the book's Example 28). These do form an A.P., with a₁ = 400 and d = 400. Using the formula, the interest at the end of 20 years is a₂₀ = 400 + 19 × 400 = Rs. 8000.
Diminishing-value depreciation is geometric: each year the value is multiplied by (1 − rate). After the first four years: 20000 × (0.94)⁴ = 20000 × 0.780749 = 15614.98. After the next six years at 8%: multiply by (0.92)⁶ = 0.606355, giving 15614.98 × 0.606355 = Rs. 9,468 (about, to the nearest rupee).
In t hours the first car travels 20t km. The second car's hourly distances are 12, 13, 14, …, an A.P., so in t hours it travels (t/2)[2 × 12 + (t − 1) × 1] = (t/2)(t + 23) km. Overtaking means equal distances: (t/2)(t + 23) = 20t. Dividing by t (t > 0) gives (t + 23)/2 = 20, so t + 23 = 40 and t = 17 hours.
⚠ Reading stated: the total work is taken as (150 workers) × (original number of days), with the workforce falling by 5 each day. Let the original plan be D days, so the work is 150D worker-days. With the falling workforce the work is done in D + 10 days, with daily workforces 150, 145, 140, …, an A.P. whose sum over D + 10 days is ((D + 10)/2)[150 + (150 − 5(D + 9))] = ((D + 10)/2)(255 − 5D). Setting this equal to 150D: (D + 10)(255 − 5D) = 300D, i.e. (D + 10)(51 − D) = 60D, i.e. 51D − D² + 510 − 10D = 60D, i.e. D² + 19D − 510 = 0. The discriminant is 361 + 2040 = 2401 = 49², so D = (−19 + 49)/2 = 15 (the negative root is discarded). The work was therefore completed in D + 10 = 25 days.
Every 5 years the level halves — a G.P. with ratio 1/2 per half-life. 20 years is 4 half-lives, so the level after 20 years is (original) × (1/2)⁴ = (original)/16. Hence the original level is 68 × 16 = 1088 microcuries.
Velocity grows by the same amount each second — an A.P. in time: v = u + at. So 20 = 4.5 + 2.5t, giving 2.5t = 15.5 and t = 6.2 seconds.
⚠ Interpretation stated (the book's wording is loose here): the temperature readings 20%, 17%, 14%, … form an A.P. with difference −3, so the fourth reading is 11%. Taking the current as proportional to the reading — 1080 mA at 20% means 1080/20 = 54 mA per percentage point — the current at the fourth measurement is 11 × 54 = 594 mA. (Equivalently, the currents 1080, 918, 756, 594 form an A.P. with difference −162 mA.) If the book meant something else by "temperature decreases", the reading of the question would change — but on the proportional reading above, 594 mA is the answer.
Let the principal be P. After 1 year the amount is P(1 + r/100); after 2 years it is P(1 + r/100)² (interest is charged on the grown amount, not just the principal); after k years it is P(1 + r/100)k. So the amounts after successive years are P(1 + r/100), P(1 + r/100)², P(1 + r/100)³, …, each obtained from the previous one by multiplying by the same factor (1 + r/100). A sequence with a constant ratio is a G.P., so the amounts form a G.P. with ratio (1 + r/100).
This chapter is about patterns that go on step by step. A sequence is a list of numbers in order — finite if it stops, infinite if it does not. When the step between terms is a fixed difference, you have an arithmetic progression (an = a₁ + (n−1)d); when the step is a fixed ratio, a geometric progression (an = a₁rn−1). Adding the terms gives a series, and the sum formulas — Sn = (n/2)(a₁ + an) for an A.P., Sn = a₁(1 − rn)/(1 − r) and S∞ = a₁/(1 − r) for a G.P. with |r| < 1 — are the workhorses of the whole chapter. The means sit between two numbers: the arithmetic mean (a + b)/2, the geometric mean ±√(ab), and the harmonic mean 2ab/(a + b), which is the reciprocal trick in disguise — a harmonic progression is just an arithmetic progression turned upside down. The arithmetico-geometric progression multiplies an A.P. by a G.P. term by term, and sigma notation with the formulas for Σk, Σk² and Σk³ lets you add whole families of series at once. The chapter closes by showing these tools at work in real life: instalments, compound interest, depreciation, population growth and falling workforces.
Revision guidance: learn the three general terms and the three sum formulas by heart first — almost every exercise question is one of them in disguise. Always check whether a question gives you the first term and the difference/ratio, or whether you must find them from two given terms. For means, remember the counting trap: inserting n means means n + 1 steps. And in series questions, try writing the general term before you reach for a formula — if it is a polynomial in k, sigma notation does the job; if it mixes an A.P. with a G.P., use the multiply-and-subtract trick.
یہ باب ان نمونوں کے بارے میں ہے جو قدم بہ قدم آگے بڑھتے ہیں۔ متوالیہ اعداد کی ایک ترتیب وار فہرست ہے — اگر رک جائے تو محدود، ورنہ لامحدود۔ جب دو متواتر رکنوں کا فرق مستقل ہو تو حسابی متوالیہ بنتا ہے (an = a₁ + (n−1)d)، اور جب نسبت مستقل ہو تو ہندسی متوالیہ (an = a₁rn−1)۔ رکنوں کو جمع کرنے سے سلسلہ بنتا ہے، اور مجموعے کے فارمولے — حسابی سلسلے کے لیے Sn = (n/2)(a₁ + an)، ہندسی سلسلے کے لیے Sn = a₁(1 − rn)/(1 − r) اور |r| < 1 کے لیے S∞ = a₁/(1 − r) — اس پورے باب کی بنیاد ہیں۔ دو اعداد کے درمیان اوسط تین طرح کی ہے: حسابی اوسط (a + b)/2، ہندسی اوسط ±√(ab)، اور توافقی اوسط 2ab/(a + b) — توافقی متوالیہ دراصل حسابی متوالیہ کا الٹا روپ ہے۔ حسابی ہندسی متوالیہ حسابی اور ہندسی متوالیہ کو رکن بہ رکن ضرب دے کر بنتا ہے، اور سگما کی علامت Σk، Σk² اور Σk³ کے فارمولوں کے ساتھ پورے سلسلوں کا مجموعہ ایک ساتھ نکالنے میں مدد دیتی ہے۔ باب کے آخر میں یہ اوزار عملی زندگی میں دکھائے گئے ہیں: قسطیں، مرکب سود، قیمت میں کمی، آبادی میں اضافہ اور کم ہوتی افرادی قوت۔
دہرائی کی رہنمائی: سب سے پہلے تین عمومی رکن اور تین مجموعے کے فارمولے زبانی یاد کریں — تقریباً ہر مشقی سوال انہی میں سے کسی ایک کا روپ ہے۔ ہمیشہ دیکھیں کہ سوال میں پہلا رکن اور فرق/نسبت دی گئی ہے یا دو رکنوں سے انہیں نکالنا ہے۔ اوسط کے سوالوں میں گنتی کی غلطی سے بچیں: n اوسط داخل کرنے کا مطلب n + 1 قدم ہے۔ اور سلسلوں کے سوالوں میں فارمولا لگانے سے پہلے عمومی رکن لکھنے کی کوشش کریں — اگر وہ k میں کثیرالرقمی ہے تو سگما کام کرے گا، اور اگر حسابی اور ہندسی کا امتزاج ہے تو ضرب و تفریق کی ترکیب استعمال کریں۔
Topics covered: Sequence · A.P. · A.M. · Arithmetic series sum · G.P. · G.M. · Geometric series · A.G.P. · H.P. and H.M. · Sigma notation · Real-life problems · Exercises 6.1–6.11
Chapter-wise tests, fresh random papers اور instant score کے لیے PunjabTest.com کا میں test section دیکھو:
PunjabTest.com — Complete Guide Book • FSc Part 1 Mathematics, Chapter 6 (New Book, Session 2026–27) • English + Urdu