JEE Main Maths: Sequences and Series questions with solutions

Q1. If Sₙ = cosⁿ θ + sinⁿ θ, then the value of 3S₄ - 2S₆ is

1. 4
2. 0
3. 1
4. 7

Answer: 1

The expression for Sₙ can be simplified using the identity for cosⁿ θ + sinⁿ θ, leading to specific values for S₄ and S₆. By substituting these values into the equation 3S₄ - 2S₆, we find that it evaluates to 1.

Q2. For a natural number n, the inequality 2^(n - 1) < nⁿ holds when

1. n < 2
2. n > 2
3. n ≥ 2
4. it never holds

Answer: n ≥ 2

For n=1, 2^0=1 is not less than 1^1=1, so it fails. For n=2, 2^1=2 < 2^2=4, and it holds for every larger n. Hence 2^(n-1) < n^n holds for n >= 2.

Q3. For any natural number n, the product n(n + 1) is always

1. an even number
2. an odd number
3. a multiple of 3
4. a multiple of 4

Answer: an even number

The product n(n + 1) consists of two consecutive integers, one of which is always even, ensuring that the product is always even.

Q4. For which positive integers n does the inequality 3ⁿ exceed n³?

1. n ≥ 2
2. n ≥ 3
3. n ≥ 4
4. n < 4

Answer: n ≥ 4

Check: n=1 (3>1) and n=2 (9>8) hold, n=3 gives 27=27 (not strictly greater) so it fails, and for n>=4 (81>64, ...) it always holds. The condition guaranteeing 3^n>n^3 for all such n is n>=4; 'n<4' is false since n=3 fails.

Q5. For which values of n does the inequality 4ⁿ/(n+1) < (2n)!/(n!)² hold, if P(n) denotes this statement?

1. n ≥ 1
2. n < 0
3. n > 0
4. n ≥ 2

Answer: n ≥ 2

The right side is C(2n,n). At n=1, 4/2 = 2 is not less than C(2,1) = 2, so it fails. For n=2, 16/3 < 6, and it continues to hold for every larger n, so the statement is true for n >= 2.

Q6. For a positive integer n, define a(n) = 1 + 1/2 + 1/3 + 1/4 +... + 1/(2ⁿ - 1). Which of the following is true?

1. a(100) is at most 100
2. a(100) is greater than 100
3. a(200) is at most 100
4. a(200) is less than 100

Answer: a(100) is at most 100

a(n)=1+1/2+...+1/(2^n -1) is approximately ln(2^n)=n ln2. So a(100) ~ 69.9 <= 100 (true), while a(200) ~ 138.6 > 100. Hence the correct statement is that a(100) is at most 100.

Q7. For which natural numbers n does the inequality 2ⁿ exceed n²?

1. n > 2
2. n > 3
3. n < 5
4. n ≥ 5

Answer: n ≥ 5

The inequality 2ⁿ exceeds n² for natural numbers n starting from 5, as for n = 5, 2⁵ = 32 and 5² = 25, and this trend continues for larger values of n due to the exponential growth of 2ⁿ compared to the polynomial growth of n².

Q8. For each natural number n, the expression n(n² - 1) is always divisible by which of the following?

1. 4
2. 6
3. 10
4. None of these

Answer: 6

The expression n(n² - 1) can be factored as n(n - 1)(n + 1), which represents the product of three consecutive integers. Since among any three consecutive integers, at least one is divisible by 2 and at least one is divisible by 3, the entire product is always divisible by 6.

Q9. If 49^λ + 16ⁿ + λ is divisible by 64 for every n ∈ N, then the greatest negative value of λ is

1. −2
2. −1
3. −3
4. −4

Answer: −1

The expression 49^λ + 16ⁿ + λ must be evaluated under modulo 64 conditions. For n being a natural number, 16ⁿ is always divisible by 64 when n ≥ 2, thus the focus is on the terms 49^λ and λ. The term 49^λ can be simplified as (49 mod 64)^λ, which is 49, and we find that for λ = -1, the expression becomes 1 + 0, satisfying the divisibility condition. Hence, the greatest negative value of λ that maintains this divisibility is -1.

Q10. If n is a natural number and n is odd, which of the following must divide n(n² - 1)?

1. 24
2. 16
3. 32
4. 19

Answer: 24

Since n is an odd natural number, n can be expressed as 2k + 1 for some integer k. The expression n(n² - 1) can be factored as n(n - 1)(n + 1), which includes three consecutive integers: n (odd), n - 1 (even), and n + 1 (even). Among these, one of the even numbers is divisible by 4, and the other is divisible by 2, ensuring that the product is divisible by 8. Additionally, since one of the three integers is odd, the product is also divisible by 3. Therefore, the overall product is divisible by 24 (8 from the even numbers and 3 from the odd number).

Q11. For every natural number n, the expression 2·7ⁿ + 3·5ⁿ − 5 is divisible by which of the following?

1. 24, for all n ∈ N
2. 21, for all n ∈ N
3. 35, for all n ∈ N
4. 50, for all n ∈ N

Answer: 24, for all n ∈ N

The expression can be simplified and analyzed for divisibility by 24, which is the product of 8 and 3. By checking the expression modulo 8 and modulo 3, we find that it consistently yields results that are divisible by both, confirming that the entire expression is divisible by 24 for all natural numbers n.

Q12. Using mathematical induction, the sum 1/(1·2·3) + 1/(2·3·4) +... + 1/((n+1)(n+2)) is equal to

1. n(n+1)/(4(n+2)(n+3))
2. n(n+3)/(4(n+1)(n+2))
3. n(n+2)/(4(n+1)(n+3))
4. None of these

Answer: n(n+3)/(4(n+1)(n+2))

The correct option is derived from the formula for the sum of the series, which can be proven using mathematical induction. By establishing a base case and showing that if the formula holds for n, it also holds for n+1, we confirm that the sum simplifies to n(n+3)/(4(n+1)(n+2)).

Q13. For each positive integer n, the number 7ⁿ − 3ⁿ is divisible by which of the following?

1. 7
2. 3
3. 4
4. 5

Answer: 4

The expression 7ⁿ − 3ⁿ can be factored and analyzed for divisibility. For any positive integer n, both 7ⁿ and 3ⁿ are odd, making their difference even, and specifically, it can be shown that it is divisible by 4.

Q14. For every integer n 1, the value of 1/(12) + 1/(23) + 1/(34) + + 1/[n(n+1)] is:

1. n/(n+1)
2. 1/(n+1)
3. 1/[n(n+1)]
4. None of these

Answer: n/(n+1)

The series can be simplified using the formula for the sum of fractions, where each term can be rewritten as a difference of two fractions, leading to a telescoping series that ultimately simplifies to n/(n+1).

Q15. Using mathematical induction, what remainder is obtained when 3^(2n) is divided by 8 for every natural number n?

1. 2
2. 3
3. 7
4. 1

Answer: 1

3^(2n) = 9^n, and 9 = 8+1, so 9^n = (8+1)^n leaves remainder 1^n = 1 when divided by 8 for every natural n.

Q16. Consider the following statements for n ∈ N: Statement 1: 1 + 2 + 3 +... + n < (1/8)(2n + 1)². Statement 2: n(n + 1)(n + 5) is divisible by 3. Which of the following is correct?

1. Only Statement 1 is true
2. Only Statement 2 is true
3. Both statements are true
4. Both statements are false

Answer: Both statements are true

Statement 1: n(n+1)/2 = (4n^2+4n)/8 < (4n^2+4n+1)/8 = (2n+1)^2/8, true for all n. Statement 2: n, n+1, n+5 cover all residues mod 3 (since n+5 ≡ n+2), so the product is divisible by 3. Both are true.

Q17. How many pairs of consecutive odd natural numbers, each greater than 10, have a sum less than 40?

1. 4
2. 6
3. 3
4. 5

Answer: 4

The pairs of consecutive odd natural numbers greater than 10 can be represented as (11, 13), (13, 15), (15, 17), and (17, 19). Their sums are 24, 28, 32, and 36 respectively, all of which are less than 40, resulting in a total of 4 valid pairs.

Q18. Evaluate the expression 2ⁿ × [1· 3· 5⋯ (2n-3)(2n-1)].

1. (2n)!/n!
2. (2n)!/2ⁿ
3. n!/(2n)!
4. None of these

Answer: (2n)!/n!

The expression represents the product of the first n odd numbers multiplied by 2 raised to the power of n, which simplifies to the factorial of 2n divided by the factorial of n, thus confirming that the correct option is (2n)!/n!.

Q19. If n is an odd positive integer, how many triples can be chosen from the set {1, 2, 3,..., n} so that the three selected numbers form an arithmetic progression?

1. (n−1)²/2
2. (n+1)²/4
3. (n+1)²/2
4. (n−1)²/4

Answer: (n−1)²/4

A 3-term AP is fixed by its first and last terms, which must have the same parity. For n = 2k+1 there are k+1 odds and k evens, giving C(k+1,2)+C(k,2) = k^2 pairs. With k = (n-1)/2 this equals (n-1)^2/4.

Q20. What remainder is obtained when 7⁹ + 9⁷ is divided by 64?

1. 0
2. 1
3. 2
4. 63

Answer: 0

Both 7⁹ and 9⁷ can be calculated modulo 64, and their individual results yield a sum that is divisible by 64, resulting in a remainder of 0.

Q21. What is the remainder when 7¹⁰³ is divided by 25?

1. 20
2. 16
3. 18
4. 15

Answer: 18

7^2 = 49 = -1 (mod 25). So 7^102 = (7^2)^51 = (-1)^51 = -1 (mod 25), and 7^103 = -7 = -7+25 = 18 (mod 25).

Q22. In the power-series expansion of (1 - 9x + 20x²)⁻¹, the coefficient of xⁿ is

1. 5ⁿ - 4ⁿ
2. 5ⁿ⁺¹ - 4ⁿ⁺¹
3. 5ⁿ⁻¹ - 4ⁿ⁻¹
4. None of these

Answer: 5ⁿ⁺¹ - 4ⁿ⁺¹

The correct option is derived from the geometric series expansion, where the function can be expressed in terms of its roots. The coefficients of the series correspond to the contributions from the roots of the polynomial, leading to the formula for the coefficients being in the form of powers of the roots, specifically resulting in the expression for the coefficient of xⁿ being 5ⁿ⁺¹ - 4ⁿ⁺¹.

Q23. Two different triples of numbers are each in arithmetic progression, and the sum of the three terms in each triple is 15. The common difference of the first triple exceeds that of the second by 1. Also, the ratio of the product of the three terms in the first triple to the product of the three terms in the second triple is 7:8. What is the ratio of the smallest term in the first triple to the smallest term in the second triple?

1. 3/4 or 11/12
2. 2/3 or 1/2
3. 2/3 or 3/4
4. None of these

Answer: 3/4 or 11/12

Sum 15 means middle term 5, so terms are 5-d,5,5+d with product 125-5d^2. The ratio (25-d1^2)/(25-d2^2)=7/8 with d1=d2+1 gives (d2,d1)=(1,2) or (-17,-16). Smallest-term ratios are 3/4 and 11/12.

Q24. Evaluate the sum of the first n terms of the series 1/(1+1²+1⁴) + 2/(1+2²+2⁴) + 3/(1+3²+3⁴) +...

1. (n²+n-1)/(2(n²+n+1))
2. (n²+n)/(2(n²+n+1))
3. (n²-n+1)/(n²+n+1)
4. (n²-n)/(2(n²+n+1))

Answer: (n²+n)/(2(n²+n+1))

k/(1+k^2+k^4)=(1/2)[1/(k^2-k+1)-1/(k^2+k+1)], a telescoping sum. The total is (1/2)[1 - 1/(n^2+n+1)] = (n^2+n)/(2(n^2+n+1)).

Q25. Find the sum of the first n terms of the series 2, 5, 14, 41,...

1. 3^(n-1)+8n-3
2. 8·3ⁿ + 4n - 8
3. 3^(n+1) + 8/3 n + 1
4. None of these

Answer: None of these

Differences are 3,9,27,... so a_k = 2 + (3^k-3)/2 = (3^k+1)/2. Summing n terms: S_n = (3^(n+1)-3)/4 + n/2, which equals none of the listed expressions, so the answer is None of these.

Q26. Let x = 1 + a + a² + a³ + ⋯ and y = 1 + b + b² + b³ + ⋯, where a and b are proper fractions. Then the value of 1 + ab + a²b² + a³b³ + ⋯ is:

1. xy/(x+y-1)
2. xy/(x-y-1)
3. xy/(x-y+1)
4. xy/(x+y+1)

Answer: xy/(x+y-1)

x=1/(1-a), y=1/(1-b), and the target sum is 1/(1-ab). With a=(x-1)/x, b=(y-1)/y, 1-ab=(x+y-1)/(xy), so 1/(1-ab)=xy/(x+y-1).

Q27. What is the value of the infinite series 1 + (1×3)/6 + (1×3×5)/(6×8) +.......... ?

1. 1
2. 0
3. ∞
4. 4

Answer: 4

Terms t_k = [1.3.5...(2k+1)]/[6.8...(2k+4)] behave like k^(-3/2), so the series converges; summing (numerically and as a hypergeometric value) gives exactly 4.

Q28. Let S1, S2, S3,..., Sn denote the sums of infinite geometric progressions whose first terms are 1, 2, 3,..., n and whose common ratios are 1/2, 1/3, 1/4,..., 1/(n+1), respectively. Then the value of S1² + S2² + S3² +... + Sn² is

1. 1/3 [n(2n+1)(4n+1)-3]
2. 1/3 [n(2n+1)(4n+1)+3]
3. 1/3 [n(2n-1)(4n+1)-3]
4. None of these

Answer: None of these

Each S_k = k/(1 - 1/(k+1)) = k+1, so the required sum is sum_{k=1}^{n} (k+1)^2 = (1/6)n(2n^2+9n+13). Checking the listed forms (e.g. at n=2 the true value is 13) none of them agree, so the answer is None of these.

Q29. If the terms a1, a2, a3,..., an form an arithmetic progression and a4 - a7 + a10 = m, then the sum of the first 13 terms of the progression is:

1. 10m
2. 12m
3. 13m
4. 15m

Answer: 13m

For an AP, a4-a7+a10 = (a+3d)-(a+6d)+(a+9d) = a+6d = a7 = m. Sum of first 13 terms = 13/2*(2a+12d) = 13(a+6d) = 13m.

Q30. If a, b, c form a geometric progression, and x and y are the arithmetic means inserted between a and b, and between b and c respectively, then the value of a/x + c/y is:

1. 0
2. 1
3. 2
4. 1/2

Answer: 2

With x=(a+b)/2 and y=(b+c)/2, a/x + c/y = 2a/(a+b) + 2c/(b+c). Using b^2=ac this simplifies to 2.

Q31. A four-term sequence has its first three terms in geometric progression and its last three terms in arithmetic progression with common difference 6. If the first term and the fourth term are equal, what is the value of the fourth term?

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

Answer: 8

The first three terms being in geometric progression means that the ratio between consecutive terms is constant, while the last three terms being in arithmetic progression with a common difference of 6 indicates that the difference between consecutive terms is fixed. Given that the first and fourth terms are equal, we can set up equations based on these properties and solve for the terms, ultimately finding that the fourth term is 8.

Q32. For the infinite series 1 + 4x + 7x² + 10x³ +..., if its sum is 35/16 and |x| < 1, then the value of x is

1. 19/7
2. 1/5
3. 1/4
4. None of these

Answer: 1/5

This arithmetic-geometric series sums to (1+2x)/(1-x)^2. Setting it equal to 35/16 gives x=1/5 or x=19/7; since |x|<1, x=1/5.

Q33. Evaluate the series 3/1² + 5/(1²+2²) + 7/(1²+2²+3²) +... taken for 11 terms:

1. 7/2
2. 11/4
3. 11/2
4. 60/11

Answer: 11/2

The kth term is (2k+1)/(sum of squares to k) = 6/(k(k+1)) = 6(1/k - 1/(k+1)). Summing 11 terms gives 6(1 - 1/12) = 6*11/12 = 11/2.

Q34. If x, y, z are greater than 1 and form a geometric progression, then the numbers 1/(1 + log x), 1/(1 + log y), and 1/(1 + log z) belong to which progression?

1. A.P.
2. H.P.
3. G.P.
4. None of these

Answer: H.P.

Since x, y, z are in geometric progression, their logarithms log x, log y, and log z are in arithmetic progression. Consequently, the reciprocals of the sums of 1 and these logarithms will form a harmonic progression.

Q35. If p, q, r form an arithmetic progression, a is the geometric mean of p and q, and b is the geometric mean of q and r, then the terms a², q², and b² are in which progression?

1. G.P.
2. A.P.
3. H.P.
4. None of these

Answer: A.P.

Since p, q, r are in arithmetic progression, the relationship between them ensures that the squares of the geometric means a and b, along with q, maintain a consistent difference, thereby forming an arithmetic progression.

Q36. Let S, P and R denote, respectively, the sum, product, and sum of the reciprocals of n terms of an increasing geometric progression. If Sⁿ = Rⁿ · P^k, then the value of k is

1. 1
2. 2
3. 3
4. None of these

Answer: 2

For an n-term GP, S/R = a^2 r^(n-1) and P^2 = a^(2n) r^(n(n-1)) = (a^2 r^(n-1))^n. Hence (S/R)^n = P^2, i.e. S^n = R^n P^2, so k = 2.

Q37. If a, b, c, d form a geometric progression, then which statement is true?

1. a + b, b + c, and c + d are in geometric progression
2. (b − c)² + (c − a)² + (d − b)² = (a − d)²
3. (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)²
4. Every statement given is correct

Answer: Every statement given is correct

With a,b,c,d=a,ar,ar^2,ar^3: (A) a+b,b+c,c+d have common ratio r -> GP; (B) (b-c)^2+(c-a)^2+(d-b)^2 = (a-d)^2 holds (e.g. 1,2,4,8 gives 49=49); (C) (a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2 holds. So every statement is correct.

Q38. Evaluate the sum of the infinite series (1)/(2) ((1)/(3)+(1)/(4))-(1)/(4) ((1)/(3²)+(1)/(4²))+(1)/(6) ((1)/(3³)+(1)/(4³))-⋯.

1. (1)/(2)log 2
2. log(3/5)
3. log(5/3)
4. (1)/(2)log(5/3)

Answer: (1)/(2)log(5/3)

The nth term is (-1)^(n+1)/(2n)[(1/3)^n+(1/4)^n]. Using sum (-1)^(n+1)x^n/n = ln(1+x), the total is (1/2)[ln(4/3)+ln(5/4)] = (1/2)ln(5/3).

Q39. If the sums of the first n terms of two arithmetic progressions are in the ratio (2n + 3): (6n + 5), then what is the ratio of their 13th terms?

1. 53: 155
2. 27: 87
3. 29: 83
4. 31: 89

Answer: 53: 155

The ratio of the m-th terms equals the sum ratio evaluated at n = 2m-1. For m = 13, n = 25: (2*25+3)/(6*25+5) = 53/155.

Q40. If a, b, c form an arithmetic progression with common difference d, then the numbers e^(1/c), e^(b/ac), and e^(1/a) are in:

1. G.P. with common ratio e^d
2. G.P. with common ratio e^(1/d)
3. G.P. with common ratio e^(d/(b²-d²))
4. A.P.

Answer: G.P. with common ratio e^(d/(b²-d²))

The numbers e^(1/c), e^(b/ac), and e^(1/a) form a geometric progression because their ratios are constant, specifically e^(d/(b²-d²)), which arises from the relationships defined by the arithmetic progression of a, b, and c.

Q41. For two positive numbers, the ratio of their harmonic mean to geometric mean is 12:13. What is the ratio of the two numbers?

1. 1:2
2. 2:3
3. 3:4
4. None of these

Answer: None of these

H/G = 2sqrt(ab)/(a+b) = 2t/(t^2+1) with t = sqrt(a/b). Setting this to 12/13 gives 6t^2 - 13t + 6 = 0, so t = 3/2 (or 2/3) and a:b = t^2 = 9:4. Since 9:4 is not listed, the answer is None of these.

Q42. If the pth, qth and rth terms of an arithmetic progression coincide with the corresponding terms of a geometric progression, and those terms are x, y and z respectively, then the value of x^(y−z), y^(z−x), z^(x−y) is

1. 0
2. 1
3. 2
4. None of these

Answer: 1

Taking logs, the exponent is (y-z)ln x+(z-x)ln y+(x-y)ln z. Writing the terms as both an AP (linear in position) and a GP (log linear in position) shows this sum is identically 0, so the product equals e^0 = 1.

Q43. If the terms a1, a2,..., a(n+1) form an arithmetic progression, then the value of 1/(a1a2) + 1/(a2a3) +... + 1/(aₙ a(n+1)) is

1. (n−1)/(a1a(n+1))
2. 1/(a1a(n+1))
3. (n+1)/(a1a(n+1))
4. n/(a1a(n+1))

Answer: n/(a1a(n+1))

With common difference d, 1/(a_i a_{i+1}) = (1/d)(1/a_i - 1/a_{i+1}). Summing telescopes to (1/d)(1/a1 - 1/a_{n+1}) = (a_{n+1}-a1)/(d a1 a_{n+1}) = n/(a1 a_{n+1}).

Q44. If the five numbers a, x, y, z, b form an arithmetic progression and x + y + z = 15, and if the same five numbers form a harmonic progression with 1/x + 1/y + 1/z = 5/3, then what are the values of a and b?

1. 2 and 8
2. 1 and 9
3. 3 and 7
4. None of these

Answer: 1 and 9

The values of a and b are determined by the properties of arithmetic and harmonic progressions. Given that the average of the arithmetic progression is equal to the middle term, and the conditions provided lead to the conclusion that a = 1 and b = 9 satisfy both the sum of the middle terms and the harmonic condition.

Q45. A ball is dropped gently from a height of 120 m. Each time it hits the ground, it rises to (4)/(5) of the height from which it fell. What is the total distance covered by the ball before it finally comes to rest?

1. 960 m
2. 1000 m
3. 1080 m
4. Infinite

Answer: 1080 m

First fall is 120 m; each subsequent up-and-down is a geometric series. Total = 120 + 2(120)(4/5)/(1-4/5) = 120 + 960 = 1080 m.

Q46. A geometric progression has an even number of terms. If the total of all its terms is five times the sum of the terms in the odd positions, then the common ratio is

1. 5
2. 1
3. 4
4. 3

Answer: 4

For a GP, the even-position terms sum to r times the odd-position sum, so total = (1+r)*S_odd. Given total = 5*S_odd, we get 1+r = 5, hence r = 4.

Q47. Evaluate the limit as n tends to infinity: (4ⁿ + 5ⁿ)^(1/n).

1. 4
2. 5
3. 5e
4. 4e/5

Answer: 5

(4^n+5^n)^(1/n) = 5*(1+(4/5)^n)^(1/n). As n->infinity, (4/5)^n->0 and the factor ->1, so the limit is 5.

Q48. Given that a1 = 1 and the sequence satisfies aₙ₊₁ = (4 + 3aₙ)/(3 + 2aₙ) for n ≥ 1, if lim_(n→∞) aₙ = a, then the value of a is

1. √2
2. −√2
3. 2
4. None of these

Answer: √2

At the limit a=(4+3a)/(3+2a) gives 2a^2=4, so a=sqrt(2). Since a1=1>0 the sequence stays positive (a2=7/5, ...), converging to the positive root sqrt(2).

Q49. Let Sₙ be the sum of the first n terms of a geometric progression with common ratio r. Then the value of (r-1) (dSₙ)/(dr) is

1. (n-1)Sₙ+nSₙ₋₁
2. (n-1)Sₙ-nSₙ₋₁
3. (n-1)Sₙ
4. None of these

Answer: (n-1)Sₙ-nSₙ₋₁

The expression for the derivative of the sum of a geometric series with respect to the common ratio captures how changes in the ratio affect the sum. The correct option reflects the relationship between the sums of the series at different terms, specifically accounting for the contributions from both the current and previous sums.

Q50. For the arithmetic progression a, a + d, a + 2d,..., a + 2nd, what is the mean deviation about the mean?

1. n(n+1)d
2. n(n+1)d/(2n+1)
3. n(n+1)d/(2n)
4. n(n-1)d/(2n+1)

Answer: n(n+1)d/(2n+1)

The AP a, a+d, ..., a+2nd has (2n+1) terms with mean a+nd. The deviations are |k-n|*d for k=0..2n, summing to 2*d*(1+2+...+n) = n(n+1)d. Dividing by (2n+1) terms gives mean deviation = n(n+1)d/(2n+1).