NCERT Solutions Class 9 Maths Ch 8: Sequences & Progressions

It lets us find any term directly, without writing all the terms before it.

• It saves time, especially for large n (for example the 1000th term).
• It shows the pattern clearly (a rule like 2n − 1 tells us the sequence is linear).
• It lets us solve problems such as “which term is equal to a given number?”.

The nth term is n².

The square numbers are 1, 4, 9, 16, 25, …, that is 1², 2², 3², 4², 5², …. So the nth term is t_n = n².

The first five terms are -3, -8, -13, -18, -23.

1. Put n = 1, 2, 3, 4, 5n = 1: 2 − 5(1) = -3n = 2: 2 − 5(2) = -8n = 3: 2 − 5(3) = -13n = 4: 2 − 5(4) = -18n = 5: 2 − 5(5) = -23

The first five terms are -1, 2, 5, 8, 11.

1. Put n = 1, 2, 3, 4, 5n = 1: 3(1) − 4 = -1n = 2: 3(2) − 4 = 2n = 3: 3(3) − 4 = 5n = 4: 3(4) − 4 = 8n = 5: 3(5) − 4 = 11

The first five terms are 2, 3, 6, 11, 18.

1. Put n = 1, 2, 3, 4, 5n = 1: 1 − 2 + 3 = 2n = 2: 4 − 4 + 3 = 3n = 3: 9 − 6 + 3 = 6n = 4: 16 − 8 + 3 = 11n = 5: 25 − 10 + 3 = 18

t₁₀ = 47 and t₁₅ = 72.

1. 10th termt₁₀ = 5(10) − 3 = 47
2. 15th termt₁₅ = 5(15) − 3 = 72

Yes, both. 97 is the 20th term and 172 is the 35th term.

1. MethodPut 5n − 3 equal to the number and solve for n. The number is a term only if n is a positive whole number.
2. For 975n − 3 = 97 ⇒ 5n = 100 ⇒ n = 20
3. For 1725n − 3 = 172 ⇒ 5n = 175 ⇒ n = 35

607 is the 122nd term.

The first five terms are −5, −2, 1, 4, 7. Yes, 52 is the 20th term.

1. First five termsEach term is 3 more than the one before.t₁ = −5,   t₂ = −2,   t₃ = 1,   t₄ = 4,   t₅ = 7
2. Is 52 a term?This is an AP with a = −5 and d = 3, so t_n = −5 + (n − 1) × 3.−5 + 3(n − 1) = 52 ⇒ 3(n − 1) = 57 ⇒ n − 1 = 19 ⇒ n = 20n is a whole number, so 52 is the 20th term.

T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81.

1. Add the three terms beforeT₄ = 4 + 2 + 1 = 7T₅ = 7 + 4 + 2 = 13T₆ = 13 + 7 + 4 = 24T₇ = 24 + 13 + 7 = 44T₈ = 44 + 24 + 13 = 81

The 10th term is 48 and the 26th term is 128.

1. IdentifyFirst term a = 3 and common difference d = 8 − 3 = 5. Use t_n = a + (n − 1)d.
2. 10th termt₁₀ = 3 + 9 × 5 = 48
3. 26th termt₂₆ = 3 + 25 × 5 = 128

−81 is the 35th term, and 0 is the 8th term, so 0 is a term.

1. Identifya = 21 and d = −3, so t_n = 21 + (n − 1)(−3) = 24 − 3n.
2. −8124 − 3n = −81 ⇒ 3n = 105 ⇒ n = 35
3. 024 − 3n = 0 ⇒ n = 8n = 8 is a positive whole number, so 0 is a term (the 8th term).

t_n = 14 − 3n. Recursive rule: t₁ = 11, t_n = t_n−1 − 3 for n ≥ 2.

1. Identifya = 11 and d = −3.
2. Explicit formulat_n = 11 + (n − 1)(−3) = 14 − 3n
3. Recursive ruleEach term is 3 less than the one before: t₁ = 11, t_n = t_n−1 − 3 (n ≥ 2).

The 29th term is 64.

1. Two equationst₃ = a + 2d = 12   and   t₅₀ = a + 49d = 106
2. Subtract47d = 94 ⇒ d = 2
3. Find aa + 2(2) = 12 ⇒ a = 8
4. 29th termt₂₉ = 8 + 28 × 2 = 64

There are 30 such numbers, and their sum is 1665.

1. The APThe 2-digit multiples of 3 are 12, 15, 18, …, 99. Here a = 12, d = 3 and the last term l = 99.
2. How manyn = 99 − 123 + 1 = 29 + 1 = 30
3. Their sumS = n2(a + l) = 302(12 + 99) = 15 × 111 = 1665

After 10 years (it is his salary in the 11th year).

1. The APa = 5,00,000 and d = 20,000.
2. When is t_n = 7,00,000?5,00,000 + (n − 1) × 20,000 = 7,00,000 ⇒ (n − 1) × 20,000 = 2,00,000 ⇒ n − 1 = 10 ⇒ n = 11
3. YearsThe salary of 7,00,000 is in the 11th year, which comes after 10 increments, that is, after 10 years.

The child uses 325 marbles.

1. The AP1, 2, 3, …, 25: a = 1, d = 1, n = 25, last term l = 25.
2. SumS₂₅ = 252(1 + 25) = 25 × 13 = 325

The 12th term is 3072.

1. Use t_n = a rⁿ⁻¹t₈ = a × 2⁷ = 128a = 192 ⇒ a = 192128 = 32
2. 12th termt₁₂ = 32 × 2¹¹ = 32 × 2048 = 3072

t₁₀ = 5¹⁰ = 9,765,625 and t_n = 5ⁿ.

1. Identifya = 5 and r = 25 ÷ 5 = 5.
2. nth termt_n = 5 × 5ⁿ⁻¹ = 5ⁿ
3. 10th termt₁₀ = 5¹⁰ = 9,765,625

730 is the 7th term.

1. Build the termst₁ = 2t₂ = 3(2) − 2 = 4t₃ = 3(4) − 2 = 10t₄ = 3(10) − 2 = 28t₅ = 3(28) − 2 = 82t₆ = 3(82) − 2 = 244t₇ = 3(244) − 2 = 730

So 730 is the 7th term. (Notice t_n − 1 = 1, 3, 9, 27, …: the terms are 3ⁿ⁻¹ + 1.)

4374 is the 8th term. Explicit: t_n = 2 · 3ⁿ⁻¹. Recursive: t₁ = 2, t_n = 3t_n−1.

1. Identifya = 2 and r = 3.
2. Explicit formulat_n = 2 · 3ⁿ⁻¹
3. Find n2 · 3ⁿ⁻¹ = 4374 ⇒ 3ⁿ⁻¹ = 2187 = 3⁷ ⇒ n − 1 = 7 ⇒ n = 8
4. Recursive formulat₁ = 2 and t_n = 3t_n−1 for n ≥ 2.

(i) 6.2208 m ≈ 6.22 m. (ii) 301.3376 m.

1. Heights form a GPAfter the kth bounce the height is 80 × (0.6)^k.

   Bounce	1	2	3	4	5
   Height (m)	48.0	28.8	17.28	10.368	6.2208

2. (i) After the 5th bounceh₅ = 80 × (0.6)⁵ = 80 × 0.07776 = 6.2208 m
3. (ii) Total distanceThe ball first falls 80 m (the 1st time it hits the ground). Then for each of the next 5 bounces it rises and falls the same height. The 6th hit comes at the end of the fall after the 5th bounce.Distance = 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208)= 80 + 2 × 110.6688 = 80 + 221.3376 = 301.3376 m

128 is the 13th term.

1. IdentifyThis is a GP with a = 2 and r = 2√2 ÷ 2 = √2 = 2^1/2.
2. Write the nth term as a power of 2t_n = 2(√2)ⁿ⁻¹ = 2 × 2^(n−1)/2 = 2^{1 + (n−1)/2}
3. Set it equal to 128 = 2⁷1 + n − 12 = 7 ⇒ n − 12 = 6 ⇒ n − 1 = 12 ⇒ n = 13

Red squares: 1, 8, 64, 512, then 4096 and 32768. Explicit: 8ⁿ. Area: (8/9)ⁿ, which tends to 0.

1. (i) Stages 0 to 3Each stage keeps 8 of the 9 equal parts of every red square, so the number of red squares is multiplied by 8.

   Stage	0	1	2	3
   Red squares	1	8	8 × 8 = 64	8 × 8 × 8 = 512

2. (ii) Stages 4 and 5Stage 4: 8⁴ = 4096   and   Stage 5: 8⁵ = 32768
3. (iii) RuleThis is a GP with ratio 8, starting at Stage 0.Explicit: s_n = 8ⁿRecursive: s₀ = 1,   s_n = 8 × s_n−1 (n ≥ 1)
4. OR: the area of the red regionEach stage keeps 8 out of 9 equal parts, so the area is multiplied by 8/9.

   Stage	0	1	2	3	4	5
   Area	11	89	6481	512729	40966561	3276859049

   Explicit: A_n = (89)ⁿRecursive: A₀ = 1,   A_n = 89 × A_n−1 (n ≥ 1)
5. What happens as n increases?Since 8/9 is less than 1, each stage has a smaller area than the one before: (8/9)ⁿ gets closer and closer to 0. The red area keeps shrinking towards zero, while the number of red squares (8ⁿ) grows without limit.

The 31st term is 178.

1. Two equationsa + 10d = 38 …(1)   and   a + 15d = 73 …(2)
2. Subtract (1) from (2)5d = 35 ⇒ d = 7
3. Find aa + 70 = 38 ⇒ a = −32
4. 31st termt₃₁ = −32 + 30 × 7 = 178

The AP is 4, 10, 16, 22, 28, …

1. Use the 7th and 5th terms(a + 6d) − (a + 4d) = 12 ⇒ 2d = 12 ⇒ d = 6
2. Use the 3rd terma + 2d = 16 ⇒ a + 12 = 16 ⇒ a = 4

So the AP is 4, 10, 16, 22, 28, … (a = 4, d = 6).

There are 128 such numbers.

1. First and lastThe first three-digit multiple of 7 is 15 × 7 = 105 (since 14 × 7 = 98). The last is 142 × 7 = 994 (since 143 × 7 = 1001).
2. Countn = 994 − 1057 + 1 = 127 + 1 = 128

There are 60 multiples of 4.

1. First and lastThe first multiple of 4 above 10 is 12 and the last below 250 is 248.
2. Countn = 248 − 124 + 1 = 59 + 1 = 60

Two GPs: −4/3, −8/3, −16/3, … (r = 2) and 4, −8, 16, −32, … (r = −2).

1. Use the 5th and 3rd termsar⁴ = 4(ar²) ⇒ r² = 4 ⇒ r = 2 or r = −2
2. Use the sum of the first two termsa + ar = a(1 + r) = −4
3. Case r = 23a = −4 ⇒ a = −43The GP is −4/3, −8/3, −16/3, −32/3, …
4. Case r = −2−a = −4 ⇒ a = 4

   The GP is 4, −8, 16, −32, …

18 + 19 + 20 + 21 + 22 and 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 (and, trivially, 100 itself).

1. Set upLet the numbers be a, a + 1, …, a + (n − 1). Their sum is n2[2a + (n − 1)] = 100, so n(2a + n − 1) = 200.
2. Try the divisors of 200n and (2a + n − 1) must have opposite parity (one even, one odd), and a must be a natural number.

   n	2a + n − 1	a	Sum
   1	200	100	100 (just one number)
   5	40	18	18 + 19 + 20 + 21 + 22
   8	25	9	9 + 10 + … + 16

3. Other nFor every other n, the two factors have the same parity, or a is not a natural number, so there are no more ways.

At the end of the 2nd hour: 120. At the end of the 4th hour: 480. At the end of the nth hour: 30 × 2ⁿ.

1. StartAt time 0 (the beginning) there are 30 bacteria. Each hour the number doubles.

   End of hour	0	1	2	3	4	n
   Bacteria	30	60	120	240	480	30 × 2n

2. RuleAfter n hours: 30 × 2ⁿ
3. Answers2nd hour: 30 × 2² = 1204th hour: 30 × 2⁴ = 480

Note: the 30 bacteria are there originally, at hour 0, so after 2 hours there are 30 × 4 = 120 and after 4 hours 30 × 16 = 480. Do not treat 30 as the count at the end of the first hour.

The first three terms are −13, −8, −3.

1. First condition(a + 3d) + (a + 7d) = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12 …(1)
2. Second condition(a + 5d) + (a + 9d) = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22 …(2)
3. Solve(2) − (1): 2d = 10 ⇒ d = 5a + 25 = 12 ⇒ a = −13
4. The first three terms−13,   −13 + 5 = −8,   −8 + 5 = −3

n = 45.

1. The conditionn(n + 1)2 > 1000 ⇒ n(n + 1) > 2000
2. Test valuesn = 44: 44 × 45 = 1980 (not enough, sum = 990)n = 45: 45 × 46 = 2070 (enough, sum = 1035)

So the smallest n is 45.

131072 is the 9th term. Explicit: a_n = 2 · 4ⁿ⁻¹. Recursive: a₁ = 2, a_n = 4a_n−1.

1. Identifya = 2 and r = 8 ÷ 2 = 4.
2. Explicit formulaa_n = 2 · 4ⁿ⁻¹
3. Find n2 · 4ⁿ⁻¹ = 131072 ⇒ 4ⁿ⁻¹ = 65536 = 4⁸ ⇒ n − 1 = 8 ⇒ n = 9
4. Recursive formulaa₁ = 2 and a_n = 4a_n−1 for n ≥ 2.

Common ratio −3/4 with terms 4/3, −1, 3/4, or common ratio −4/3 with terms 3/4, −1, 4/3.

1. Let the terms be a, ar, ar²Product: a · ar · ar² = (ar)³ = −1, so ar = −1 and a = −1/r.
2. Use the suma(1 + r + r²) = 1312 ⇒ −1r(1 + r + r²) = 1312−12(1 + r + r²) = 13r ⇒ 12r² + 25r + 12 = 0
3. Solver = −25 ± √(625 − 576)24 = −25 ± 724r = −34   or   r = −43
4. The termsIf r = −3/4, then a = 4/3: 4/3, −1, 3/4. If r = −4/3, then a = 3/4: 3/4, −1, 4/3.

y² = xz, so x, y, z are in GP.

1. Write the termsx = ar³, y = ar⁹, z = ar¹⁵.
2. Compare y² and xzy² = (ar⁹)² = a²r¹⁸xz = (ar³)(ar¹⁵) = a²r¹⁸
3. ConclusionSince y² = xz, we have y ÷ x = z ÷ y, so x, y, z are in GP (with common ratio r⁶).

The terms are 2, 6, 18 or 18, 6, 2.

1. Equationsa(1 + r + r²) = 26 …(1)a²(1 + r² + r⁴) = 364 …(2)
2. Divide (2) by (1) squared1 + r² + r⁴(1 + r + r²)² = 364676 = 713Use 1 + r² + r⁴ = (1 + r + r²)(1 − r + r²):1 − r + r²1 + r + r² = 713
3. Solve13 − 13r + 13r² = 7 + 7r + 7r² ⇒ 6r² − 20r + 6 = 0 ⇒ 3r² − 10r + 3 = 0r = 3   or   r = 13
4. Find aIf r = 3: 13a = 26, so a = 2 and the terms are 2, 6, 18.

   If r = 1/3: a(1 + 1/3 + 1/9) = 26 gives a = 18, and the terms are 18, 6, 2.

Check: 2 + 6 + 18 = 26 and 4 + 36 + 324 = 364.

The terms are 1, 2, 4, 8, 16, 32, 64, 128. Recursive: P_n = 2P_n−1. Explicit: P_n = 2ⁿ⁻¹.

1. The valuesP₃ = 1 + 2 + 1 = 4P₄ = 1 + 2 + 4 + 1 = 8P₅ = 1 + 2 + 4 + 8 + 1 = 16

   and in the same way P₆ = 32, P₇ = 64, P₈ = 128.

2. A simpler recursive formulaEvery term is double the term before: P₁ = 1, P_n = 2P_n−1 (n ≥ 2).
3. Why it worksP_n+1 − P_n = P_n (the new sum has one extra term, P_n), so P_n+1 = 2P_n.
4. Explicit formulaThis is a GP with a = 1 and r = 2: P_n = 2ⁿ⁻¹.

The terms are 1, 2, 3, 5, 8, 13, 21, 34. This is the Fibonacci sequence.

1. The valuesW₃ = W₁ + 2 = 3W₄ = W₁ + W₂ + 2 = 5W₅ = 1 + 2 + 3 + 2 = 8W₆ = 1 + 2 + 3 + 5 + 2 = 13W₇ = 1 + 2 + 3 + 5 + 8 + 2 = 21W₈ = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34
2. Recognise the sequence1, 2, 3, 5, 8, 13, 21, 34: each term from the third onwards is the sum of the two before it (3 = 1 + 2, 5 = 2 + 3, 8 = 3 + 5, …). This is the Fibonacci sequence.