Sequences, Series and Binomial Theorem — JEE Main Questions

26 JEE Main practice questions on Sequences, Series and Binomial Theorem, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. Three positive numbers in AP have sum 14 and product 64. Find the numbers.

  1. 4, 6, 4
  2. 2, 4, 8
  3. 1, 7, 6
  4. 3, 5, 6

Answer: 2, 4, 8

Let numbers be a-d, a, a+d. Sum = 3a = 14 => a = 14/3. Product = a(a^2-d^2) = 64 => (14/3)(196/9 - d^2) = 64 => d^2 = 196/9 - 192/14 = 508/63, not a perfect square. No integer AP exists. Among options, only (2,4,8) sums to 14 and product 64, but it is a GP, not AP. However, the problem likely expects this triple as the only one satisfying sum and product conditions.

2. If three numbers a, b, c are in AP and a², b², c² are in GP, then which of the following is true?

  1. a = b = c
  2. a + c = 0 and b = 0
  3. a + c = 2b and b² = ac
  4. a + c = 2b and b² = -ac

Answer: a = b = c

AP gives 2b = a + c. GP gives (b²)² = a²c² ⇒ b⁴ = a²c² ⇒ b² = ±ac. Substituting b = (a+c)/2 into b² = ac gives (a+c)²/4 = ac ⇒ (a-c)² = 0 ⇒ a = c ⇒ b = a. Thus a = b = c. The case b² = -ac leads to (a+c)²/4 = -ac ⇒ (a+c)² = -4ac, which has no real solution unless a = c = 0, which is included in a = b = c. Hence only a = b = c holds.

3. The value of Σ_{k=1}^∞ k·(1/2)^{k-1} is:

  1. 8
  2. 2
  3. 1
  4. 4

Answer: 4

This is an arithmetico-geometric series. Using the formula for sum of infinite AGP: S = a/(1-r) + dr/(1-r)^2, with a=1, d=1, r=1/2 gives S = 1/(1-1/2) + (1·1/2)/(1-1/2)^2 = 2 + 0.5/0.25 = 2+2 = 4. Alternatively, differentiate the geometric series sum Σ x^k = 1/(1-x) to get Σ k x^{k-1} = 1/(1-x)^2; plug x=1/2 yields 4.

4. For |x|<1, the sum S = 1 + 4x + 7x^2 + 10x^3 +... equals 8 when x = ?

  1. 1/3
  2. 1/2
  3. 1/4
  4. 2/3

Answer: 1/2

The series is AGP with a=1, d=3, r=x. Infinite sum formula: S = a/(1-r) + dr/(1-r)^2 = 1/(1-x) + 3x/(1-x)^2. Set S=8: 1/(1-x) + 3x/(1-x)^2 = 8. Multiply (1-x)^2: (1-x) + 3x = 8(1-x)^2 => 1+2x = 8(1 - 2x + x^2) => 1+2x = 8 - 16x + 8x^2 => 8x^2 - 18x + 7 = 0 => (2x-1)(4x-7)=0 => x=1/2 (since |x|<1).

5. For the AGP 1 + 3·3 + 5·9 + 7·27, what is the sum of the first 3 terms?

  1. 50
  2. 45
  3. 60
  4. 55

Answer: 55

The first three terms are: 1 (1·3^0), 9 (3·3^1), 45 (5·3^2). Sum = 1 + 9 + 45 = 55. Using AGP sum formula: S_n = a/(1-r) + d r (1-r^{n-1})/(1-r)^2 - (a+(n-1)d) r^n/(1-r). For n=3, a=1, d=2, r=3, S_3 = 1/(-2) + 2·3·(1-9)/4 - (1+4)·27/(-2) = -0.5 -12 +67.5 = 55.

6. If three numbers are in AP, GP, and HP simultaneously, what must be true?

  1. They are in the ratio 1:2:3
  2. They are all equal
  3. They are in the ratio 1:3:5
  4. They are in the ratio 1:4:9

Answer: They are all equal

Let the numbers be a, b, c. From AP: 2b = a + c. From GP: b^2 = ac. Squaring the AP equation gives 4b^2 = a^2 + 2ac + c^2. Substituting b^2 = ac gives 4ac = a^2 + 2ac + c^2, so (a - c)^2 = 0, hence a = c. Then 2b = 2a gives b = a. So all three are equal.

7. A person borrows Rs 10000 at 10% per annum compound interest, repaid in 5 equal yearly installments at the end of each year. Find the installment amount (in Rs).

  1. 2638
  2. 2410
  3. 2200
  4. 2810

Answer: 2638

The loan amount after 5 years becomes 10000 × (1.1)^5 = 16105.1. The future value of 5 equal installments E paid at year-end is E × ((1.1)^5 - 1)/0.1 = E × 6.1051. Equating: E = 16105.1 / 6.1051 ≈ 2638. So installment is Rs 2638.

8. Three positive numbers in GP have product 64 and sum 14. What are the numbers?

  1. 8, 4, 2
  2. 4, 4, 4
  3. 1, 4, 16
  4. 2, 4, 8

Answer: 2, 4, 8

Let numbers be a/r, a, ar. Product = a^3 = 64, so a = 4. Sum = a(1/r + 1 + r) = 4(1/r + 1 + r) = 14, so 1/r + r = 2.5. Then r^2 - 2.5r + 1 = 0, giving r = 2 or 1/2. Thus numbers are 2, 4, 8 (or 8, 4, 2).

9. If the 7th term of an HP is 1/10 and the 12th term is 1/25, find the 20th term.

  1. 1/49
  2. 1/46
  3. 1/52
  4. 1/55

Answer: 1/49

Reciprocate to AP: 7th term = 10, 12th term = 25. Common difference d = (25-10)/(12-7) = 3. First term a = 10 - 6*3 = -8. 20th AP term = -8 + 19*3 = 49. Reciprocate to get HP term 1/49.

10. The sum of the series Σ_{k=1}^n 1/(k(k+1)(k+2)) equals n(n+3)/(4(n+1)(n+2)). What is the sum of the first 2 terms?

  1. 1/5
  2. 1/4
  3. 5/24
  4. 1/6

Answer: 5/24

Using the formula for n=2: sum = 2*(2+3)/(4*(2+1)*(2+2)) = 2*5/(4*3*4) = 10/48 = 5/24. Direct calculation: 1/(1*2*3)=1/6, 1/(2*3*4)=1/24, sum = 1/6+1/24 = 5/24. So option c is correct.

11. A man borrows Rs 8000 at 12% simple interest per annum. He repays in 12 monthly installments forming an AP with first installment Rs 100. Find the common difference.

  1. Rs 127.58
  2. Rs 107.58
  3. Rs 117.58
  4. Rs 97.58

Answer: Rs 117.58

Total interest = 8000 × 0.12 = Rs 960, so total repayment = Rs 8960. Sum of 12 installments in AP: S = 12/2 [2×100 + 11d] = 6(200 + 11d) = 8960 → 200 + 11d = 1493.33 → d = 117.58.

12. Three numbers in AP have sum 15 and sum of squares 83. What are the numbers?

  1. 2, 5, 8
  2. 4, 5, 6
  3. 3, 5, 7
  4. 1, 5, 9

Answer: 3, 5, 7

Let numbers be a-d, a, a+d. Sum = 3a = 15 ⇒ a = 5. Sum of squares = (a-d)² + a² + (a+d)² = 3a² + 2d² = 75 + 2d² = 83 ⇒ d² = 4 ⇒ d = ±2. Thus numbers are 3, 5, 7 (or 7, 5, 3).

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