Exercise 8.2 answers: Sequences and Series

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1. First term a = 5/2. Common ratio r = (5/4) ÷ (5/2) = 1/2.
2. For a G.P., the nth term is a × r^(n−1).
3. nth term = (5/2) × (1/2)^(n−1) = 5/2ⁿ
4. 20th term = 5/2²⁰ = 5/1048576

Answernth term = 5/2ⁿ; 20th term = 5/1048576

Watch this explained “The rule, and its off-by-one”, 3:31 into Reaching any term without walking through the earlier ones

Open NCERT p. 145Matches NCERT’s answer

1. 8th term = a × r⁷ = 192, with r = 2, so a × 128 = 192, giving a = 3/2.
2. 12th term = a × r¹¹ = (3/2) × 2048 = 3072

Answer12th term = 3072

Watch this explained “Two terms from the middle”, 9:09 into Reaching any term without walking through the earlier ones

Open NCERT p. 145One way to think about it

1. Let the G.P. have first term a and common ratio r.
2. 5th term: p = a·r⁴. 8th term: q = a·r⁷. 11th term: s = a·r¹⁰.
3. Square the middle one: q² = a²·r¹⁴.
4. Multiply the outer two: p·s = (a·r⁴)(a·r¹⁰) = a²·r¹⁴.
5. Both sides equal a²·r¹⁴, so q² = ps.

In shortq² = ps, since both equal a²r¹⁴.

Watch this explained “Equal gaps, signs, and a whole run”, 10:30 into Reaching any term without walking through the earlier ones

Open NCERT p. 145Matches NCERT’s answer

1. Let the first term be a = −3 and common ratio be r.
2. 4th term = a·r³. 2nd term = a·r, so (2nd term)² = a²·r².
3. Condition: a·r³ = a²·r², so r = a (dividing both sides by a·r², since a ≠ 0, r ≠ 0).
4. So r = a = −3.
5. 7th term = a·r⁶ = (−3) × (−3)⁶ = (−3) × 729 = −2187

Answer7th term = −2187

Watch this explained “The rule, and its off-by-one”, 3:31 into Reaching any term without walking through the earlier ones

Open NCERT p. 145Matches NCERT’s answer

(a) 2, 2√2, 4,... is 128 ?

1. a = 2, r = 2√2/2 = √2.
2. nth term: 2 × (√2)^(n−1) = 128, so (√2)^(n−1) = 64 = 2⁶.
3. Since √2 = 2^(1/2), (√2)^(n−1) = 2^((n−1)/2), so (n−1)/2 = 6.
4. n − 1 = 12, so n = 13

Answer13th term

(b) √3, 3, 3√3,... is 729 ?

1. a = √3, r = 3/√3 = √3.
2. nth term: √3 × (√3)^(n−1) = (√3)ⁿ = 729.
3. Since √3 = 3^(1/2), (√3)ⁿ = 3^(n/2) = 3⁶ (as 729 = 3⁶), so n/2 = 6.
4. n = 12

Answer12th term

(c) 1/3, 1/9, 1/27,... is 1/19683 ?

1. a = 1/3, r = 1/3.
2. nth term: (1/3)ⁿ = 1/19683.
3. 19683 = 3⁹, so (1/3)ⁿ = (1/3)⁹.
4. n = 9

Answer9th term

Watch this explained “When the unknown is the exponent”, 6:58 into Reaching any term without walking through the earlier ones

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1. For three numbers in a G.P., the middle one squared equals the product of the outer two: x² = (−2/7) × (−7/2).
2. (−2/7) × (−7/2) = 1
3. So x² = 1, giving x = 1 or x = −1

Answerx = 1 or x = −1

Watch this explained “What only the test settles”, 11:01 into A constant ratio between neighbours is the entire definition

Open NCERT p. 145Matches NCERT’s answer

1. First term a = 0.15 = 3/20. Common ratio r = 0.015/0.15 = 1/10.
2. For n terms of a G.P., sum = a(1 − rⁿ)/(1 − r).
3. S₂₀ = (3/20) × (1 − (1/10)²⁰) / (1 − 1/10) = (3/20) × (1 − (1/10)²⁰) × (10/9)
4. S₂₀ = (1/6) × (1 − (1/10)²⁰)

AnswerS₂₀ = (1/6)(1 − (1/10)²⁰)

Watch this explained “Forwards: total a stated run”, 7:42 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 145Matches NCERT’s answer

1. First term a = √7. Common ratio r = √21 ÷ √7 = √3. (Check: √7 × √3 × √3 = 3√7, which is the third term.)
2. Since r = √3 is more than 1, use the sum formula Sₙ = a(rⁿ − 1)/(r − 1).
3. Sₙ = √7((√3)ⁿ − 1)/(√3 − 1)
4. Multiply the top and the bottom by (√3 + 1). The bottom becomes (√3 − 1)(√3 + 1) = 3 − 1 = 2.
5. Sₙ = √7(√3 + 1)((√3)ⁿ − 1)/2

AnswerSₙ = √7(√3 + 1)((√3)ⁿ − 1)/2, that is, √7(√3 + 1)(3^(n/2) − 1)/2

Watch this explained “Two faces, one formula”, 6:37 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 145Matches NCERT’s answer

1. Each term is the one before it multiplied by − a (1 × (− a) = − a, (− a) × (− a) = a², …), so this is a G.P. with first term 1 and common ratio − a.
2. Sum of n terms of a G.P. = (first term) × (1 − ratioⁿ) / (1 − ratio). This works whenever the ratio is not 1.
3. Here the ratio − a is not 1, because a ≠ − 1 — that is exactly why the question gives this condition.
4. Sum = 1 × [1 − (− a)ⁿ] / [1 − (− a)] = [1 − (− a)ⁿ] / (1 + a).

Answer[1 − (− a)ⁿ] / (1 + a)

Watch this explained “The small print is the derivation”, 12:33 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 145Matches NCERT’s answer

1. This is a G.P. with first term x³ and common ratio x².
2. Use the G.P. sum formula: sum = a(rⁿ − 1)/(r − 1).
3. Put a = x³ and r = x².
4. Sum = x³(x²ⁿ − 1) / (x² − 1).

Answerx³(x²ⁿ − 1) / (x² − 1)

Watch this explained “The small print is the derivation”, 12:33 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 145Matches NCERT’s answer

1. The sigma sign means: put k = 1, 2, 3, … up to 11 into (2 + 3ᵏ) and add all the results.
2. Split the sum into two parts: 2 added 11 times, and 3ᵏ added for k = 1 to 11.
3. 2 added 11 times = 22.
4. 3, 3², 3³, … 3¹¹ is a G.P. with first term 3 and ratio 3, so its sum of 11 terms = 3(3¹¹ − 1)/(3 − 1).
5. That G.P. sum works out to 265719.
6. Total = 22 + 265719 = 265741.

Answer265741

Watch this explained “Forwards: total a stated run”, 7:42 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 145Checked by computer

1. Write the three terms centred on the middle term: a/r, a, ar.
2. Their product is (a/r) × a × (ar) = a³. This equals 1, so a = 1.
3. Their sum is 1/r + 1 + r = 39/10.
4. Multiply every term by 10r: 10 + 10r + 10r² = 39r, so 10r² − 29r + 10 = 0.
5. Factorise: (5r − 2)(2r − 5) = 0, so r = 5/2 or r = 2/5.
6. With r = 5/2 the terms a/r, a, ar are 2/5, 1, 5/2. Check: 2/5 + 1 + 5/2 = 4/10 + 10/10 + 25/10 = 39/10, and 2/5 × 1 × 5/2 = 1.
7. With r = 2/5 you get the same three numbers in reverse order, 5/2, 1, 2/5. Both answers are correct.

AnswerCommon ratio 5/2, terms 2/5, 1, 5/2 (or common ratio 2/5, terms 5/2, 1, 2/5).

Watch this explained “A total and a product together”, 11:21 into Subtracting a scaled copy of the total to collapse it to two terms

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1. This G.P. has first term a = 3 and common ratio r = 3.
2. Sum of n terms = a(rⁿ − 1)/(r − 1) = 3(3ⁿ − 1)/2.
3. Set this equal to 120: 3(3ⁿ − 1)/2 = 120, so 3ⁿ = 81.
4. 81 = 3⁴, so n = 4.

Answer4 terms

Watch this explained “Backwards: the run length is the unknown”, 8:45 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 145Matches NCERT’s answer

1. Let the G.P. have first term a and ratio r. The first three terms are a, ar, ar²; the next three are ar³, ar⁴, ar⁵.
2. The next three terms are the first three shifted 3 steps along, so their sum is r³ times the sum of the first three terms.
3. 128 = r³ × 16, so r³ = 8, giving r = 2.
4. Sum of first three terms: a(1 + r + r²) = 16, so 7a = 16, giving a = 16/7.
5. Sum to n terms = a(rⁿ − 1)/(r − 1) = (16/7)(2ⁿ − 1)/1 = 16(2ⁿ − 1)/7.

AnswerFirst term = 16/7, common ratio = 2, sum to n terms = 16(2ⁿ − 1)/7.

Watch this explained “Dividing leaves the gap”, 8:01 into Reaching any term without walking through the earlier ones

Open NCERT p. 145Checked by computer

1. The 7th term of a G.P. is a·r⁶. With a = 729: 729·r⁶ = 64, so r⁶ = 64/729 = (2/3)⁶.
2. An even power hides the sign, so r = 2/3 or r = − 2/3. The usual answer takes r = 2/3.
3. With r = 2/3: S₇ = a(1 − r⁷)/(1 − r) = 729 × (1 − 128/2187) ÷ (1/3).
4. 1 − 128/2187 = 2059/2187, and 729 × 2059/2187 = 2059/3. Dividing by 1/3 means multiplying by 3, so S₇ = 2059.
5. (If r = − 2/3 is taken instead, S₇ = 729 × (1 + 128/2187) ÷ (5/3) = 463.)

AnswerS₇ = 2059 (with r = 2/3; the ratio r = − 2/3 also fits and gives S₇ = 463)

Watch this explained “Forwards: total a stated run”, 7:42 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 146Checked by computer

1. Let the first term be a and the common ratio r. The third term is ar² and the fifth is ar⁴.
2. Fifth term = 4 × third term: ar⁴ = 4ar². No term of a G.P. is 0, so divide by ar²: r² = 4, so r = 2 or r = − 2.
3. Sum of the first two terms: a + ar = a(1 + r) = − 4.
4. Take r = − 2: a(1 − 2) = − 4, so − a = − 4 and a = 4. The G.P. is 4, − 8, 16, − 32, 64, …
5. Check: 4 + (− 8) = − 4, and the fifth term 64 = 4 × 16, four times the third term.
6. Take r = 2 instead: 3a = − 4, so a = − 4/3, giving − 4/3, − 8/3, − 16/3, … This also fits. The question asks for a G.P., so either one is a correct answer.

AnswerOne such G.P.: first term 4, common ratio − 2, i.e. 4, − 8, 16, − 32, … (another is − 4/3, − 8/3, − 16/3, … with ratio 2).

Watch this explained “Dividing leaves the gap”, 8:01 into Reaching any term without walking through the earlier ones

Open NCERT p. 146One way to think about it

1. Let the G.P. have first term a and ratio r. Then x = ar³, y = ar⁹, z = ar¹⁵.
2. The gap from the 4th term to the 10th is 6 steps, and from the 10th to the 16th is also 6 steps — equal gaps.
3. y/x = ar⁹/ar³ = r⁶, and z/y = ar¹⁵/ar⁹ = r⁶.
4. Since y/x = z/y, x, y, z have a constant ratio, so they are in G.P.

In shorty/x = z/y = r⁶, so x, y, z are in G.P.

Watch this explained “Equal gaps, signs, and a whole run”, 10:30 into Reaching any term without walking through the earlier ones

Open NCERT p. 146Matches NCERT’s answer

1. Write each term as 8 times a string of 1s: 8 = 8 × 1, 88 = 8 × 11, 888 = 8 × 111, and so on.
2. A string of k ones equals (10^k − 1)/9. For example 111 = 999/9 = (10³ − 1)/9.
3. So S_n = (8/9)[(10 − 1) + (10² − 1) + … + (10ⁿ − 1)] = (8/9)[(10 + 10² + … + 10ⁿ) − n].
4. 10 + 10² + … + 10ⁿ is a G.P. with first term 10 and ratio 10, so its sum is 10(10ⁿ − 1)/(10 − 1) = 10(10ⁿ − 1)/9.
5. S_n = (8/9)[10(10ⁿ − 1)/9 − n] = (8/81)[10(10ⁿ − 1) − 9n] = 8(10ⁿ⁺¹ − 9n − 10)/81.
6. Check with n = 2: 8(1000 − 18 − 10)/81 = 8 × 972/81 = 8 × 12 = 96, and 8 + 88 = 96.

AnswerS_n = 8(10ⁿ⁺¹ − 9n − 10)/81

Watch this explained “Check first, convert second”, 9:56 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 146Matches NCERT’s answer

1. Multiply the terms standing in the same position: 2×128, 4×32, 8×8, 16×2, 32×(1/2).
2. These products are 256, 128, 64, 32, 16.
3. Adding them: 256 + 128 + 64 + 32 + 16 = 496.

Answer496

Watch this explained “Forwards: total a stated run”, 7:42 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 146Matches NCERT’s answer

1. The k-th term (counting from k = 0) of the first sequence is a·rᵏ, and of the second is A·Rᵏ.
2. Their product is (aA)·(rR)ᵏ.
3. The ratio between consecutive products, term (k+1) over term k, is (aA)(rR)ᵏ⁺¹ / (aA)(rR)ᵏ = rR — the same for every k.
4. A constant ratio between consecutive terms means the product sequence is itself a G.P., with common ratio rR.

AnswerThe product sequence is a G.P. with common ratio rR.

Watch this explained “What only the test settles”, 11:01 into A constant ratio between neighbours is the entire definition

Open NCERT p. 146Matches NCERT’s answer

1. Let the four numbers be a, ar, ar², ar³.
2. Third term greater than first by 9: ar² − a = 9, i.e. a(r² − 1) = 9.
3. Second term greater than fourth by 18: ar − ar³ = 18, i.e. − ar(r² − 1) = 18.
4. Dividing the second equation by the first: − r = 18/9 = 2, so r = − 2.
5. Substitute back: a(4 − 1) = 9, so a = 3.
6. The four numbers are 3, − 6, 12, − 24.

Answer3, − 6, 12, − 24

Watch this explained “Two numbers fix the whole thing”, 7:55 into A constant ratio between neighbours is the entire definition

Open NCERT p. 146One way to think about it

1. The letter r is already a position in the question, so call the first term A and the common ratio R. Then a = AR^p−1, b = AR^q−1, c = AR^r−1.
2. So a^q−r b^r−p c^p−q = A^{(q−r)+(r−p)+(p−q)} × R^{(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q)}.
3. Power of A: (q − r) + (r − p) + (p − q) = 0.
4. Power of R, expanding each bracket: (p−1)(q−r) = pq − pr − q + r; (q−1)(r−p) = qr − pq − r + p; (r−1)(p−q) = pr − qr − p + q.
5. Adding these three, every term cancels with another (pq with − pq, pr with − pr, qr with − qr, p with − p, q with − q, r with − r), so the power of R is 0.
6. So the product is A⁰ × R⁰ = 1 × 1 = 1.

In shorta^q−r b^r−p c^p−q = A⁰R⁰ = 1, since both powers add up to 0.

Watch this explained “The rule, and its off-by-one”, 3:31 into Reaching any term without walking through the earlier ones

Open NCERT p. 146One way to think about it

1. Let the common ratio be r. The terms are a, ar, ar², …, arⁿ⁻¹, and the nth term is b = arⁿ⁻¹.
2. P = a × ar × ar² × … × arⁿ⁻¹. Write the same product backwards: P = arⁿ⁻¹ × arⁿ⁻² × … × a.
3. Multiply the two lines factor by factor. The kth factor forwards is ar^k−1 and the kth backwards is ar^n−k, so their product is a²rⁿ⁻¹ = a × arⁿ⁻¹ = ab — the same for every k.
4. There are n such products, so P × P = (ab)ⁿ, that is, P² = (ab)ⁿ.

In shortP² = (ab)ⁿ, because each of the n pairs multiplies to ab.

Watch this explained “Equal gaps, signs, and a whole run”, 10:30 into Reaching any term without walking through the earlier ones

Open NCERT p. 146One way to think about it

1. Let the G.P. have first term a and ratio r. Sum of the first n terms: Sₙ = a(rⁿ − 1)/(r − 1).
2. The terms from position (n+1) to (2n) are the same n terms as the first block, but each multiplied by rⁿ (shifted n steps along).
3. So their sum is rⁿ times Sₙ.
4. Ratio = Sₙ / (rⁿ·Sₙ) = 1/rⁿ.

In shortThe ratio is 1/rⁿ, because the second block of n terms is the first block scaled by rⁿ.

Watch this explained “Dividing leaves the gap”, 8:01 into Reaching any term without walking through the earlier ones

Open NCERT p. 146One way to think about it

1. Since a, b, c, d are in G.P. with ratio r: b = ar, c = ar², d = ar³.
2. Then ab = a²r, bc = a²r³, cd = a²r⁵, so ab + bc + cd = a²r(1 + r² + r⁴).
3. Also a² + b² + c² = a²(1 + r² + r⁴), and b² + c² + d² = a²r²(1 + r² + r⁴).
4. Multiplying: (a²+b²+c²)(b²+c²+d²) = a⁴r²(1 + r² + r⁴)².
5. Squaring the right side: (ab+bc+cd)² = [a²r(1+r²+r⁴)]² = a⁴r²(1 + r² + r⁴)².
6. Both sides are equal.

In shortBoth sides equal a⁴r²(1 + r² + r⁴)², so the identity holds.

Watch this explained “Two numbers fix the whole thing”, 7:55 into A constant ratio between neighbours is the entire definition

Open NCERT p. 146Matches NCERT’s answer

1. The sequence is 3, x, y, 81 — a G.P. of 4 terms, with first term 3 and 4th term 81.
2. 4th term = first term × r³, so 81 = 3 × r³, giving r³ = 27, so r = 3.
3. x = 3 × r = 9, and y = 9 × r = 27.

AnswerThe two numbers are 9 and 27, giving the G.P. 3, 9, 27, 81.

Watch this explained “How many answers”, 8:42 into The number that sits between two others multiplicatively

Open NCERT p. 146Matches NCERT’s answer

1. Given: (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) is the geometric mean between a and b.
2. The geometric mean of a and b (positive, a ≠ b) is √(ab) = a^1/2b^1/2.
3. We need (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) = a^1/2b^1/2. Cross-multiply: aⁿ⁺¹ + bⁿ⁺¹ = a^n+1/2b^1/2 + a^1/2b^n+1/2.
4. Collect the a-powers on one side: aⁿ⁺¹ − a^n+1/2b^1/2 = a^1/2b^n+1/2 − bⁿ⁺¹, i.e. a^n+1/2(a^1/2 − b^1/2) = b^n+1/2(a^1/2 − b^1/2).
5. Since a ≠ b, a^1/2 − b^1/2 is not 0, so divide by it: a^n+1/2 = b^n+1/2, i.e. (a/b)^n+1/2 = 1.
6. a/b is a positive number other than 1, so its power can equal 1 only when the power is 0: n + 1/2 = 0, so n = − 1/2.
7. Check: with n = − 1/2 the top is √a + √b and the bottom is 1/√a + 1/√b = (√a + √b)/√(ab), so the fraction is √(ab).

Answern = − 1/2

Watch this explained “Solving for the exponent”, 11:53 into The number that sits between two others multiplicatively

Open NCERT p. 146One way to think about it

1. Let the two numbers be p and q
2. Given: p + q = 6√(pq)
3. Let g = √(pq), so pq = g²
4. Then: p + q = 6g
5. p and q are roots of the equation t² − 6gt + g² = 0
6. Using the quadratic formula: t = (6g ± √(36g² − 4g²))/2
7. Simplifying: t = (6g ± √(32g²))/2 = (6g ± 4g√2)/2
8. t = g(3 ± 2√2)
9. Therefore, the two numbers are g(3 + 2√2) and g(3 − 2√2)
10. The ratio is (3 + 2√2) : (3 − 2√2) ✓

In shortProved

Watch this explained “The same fact, other clothes”, 9:08 into Why one of the two means can never overtake the other

Open NCERT p. 146One way to think about it

1. Let the two positive numbers be p and q. A = (p+q)/2, so p + q = 2A. G = √(pq), so pq = G².
2. (p − q)² = (p + q)² − 4pq = 4A² − 4G² = 4(A+G)(A−G).
3. So p − q = 2√((A+G)(A−G)) (taking p ≥ q).
4. Adding p + q = 2A and p − q = 2√((A+G)(A−G)): p = A + √((A+G)(A−G)).
5. Subtracting instead: q = A − √((A+G)(A−G)).

In shortThe numbers are A + √((A+G)(A−G)) and A − √((A+G)(A−G)).

Watch this explained “The same fact, other clothes”, 9:08 into Why one of the two means can never overtake the other

Open NCERT p. 146Matches NCERT’s answer

1. Originally, before any hour has passed, there are 30 bacteria. Each hour the number is multiplied by 2.
2. End of 1st hour: 30 × 2 = 60. End of 2nd hour: 60 × 2 = 120, which is 30 × 2².
3. After n hours the 30 has been doubled n times, so the count is 30 × 2ⁿ. (In the G.P. 30, 60, 120, … with first term 30 and ratio 2, the end of the nth hour is the (n + 1)th term, and 30 × 2^(n+1)−1 = 30 × 2ⁿ.)
4. End of 4th hour: 30 × 2⁴ = 30 × 16 = 480.

AnswerEnd of 2nd hour: 120; end of 4th hour: 480; end of nth hour: 30 × 2ⁿ.

Watch this explained “Count the steps, not the terms”, 1:00 into Reaching any term without walking through the earlier ones

Open NCERT p. 146Matches NCERT’s answer

1. Compounding at 10% every year means each year's amount is the previous year's amount × 1.1 — that is a G.P. with first term 500 and common ratio 1.1.
2. After 10 years, the amount is the 11th term of this G.P.: 500 × (1.1)¹⁰.
3. Working this out exactly gives 25937424601/20000000, which is about ₹1296.87.

Answer≈ ₹1296.87 (exactly 25937424601/20000000)

Watch this explained “The rule, and its off-by-one”, 3:31 into Reaching any term without walking through the earlier ones

Open NCERT p. 146Matches NCERT’s answer

1. If the roots are α and β, the A.M. is (α+β)/2 = 8, so α+β = 16.
2. The G.M. is √(αβ) = 5, so αβ = 25.
3. A quadratic with roots α, β can be written x² − (sum of roots)x + (product of roots) = 0.
4. Substituting the values: x² − 16x + 25 = 0.

Answerx² − 16x + 25 = 0

Watch this explained “The middles carry the pair”, 6:09 into Why one of the two means can never overtake the other