JEE Main Sequences and Series questions
58 Mathematics questions on Sequences and Series as asked in JEE Main, newest first. 33 come with a worked solution.
- Let f(x) be a function such that f(x + y) = f(x) · f(y) for all x, y ∈ N. If f(1) = 3 and sum_k=1^n f(k) = 3279, then the value of n is
2023
- For three positive integers p, q, r, x^pq^2 = y^qr = z^p^2r and r = pq + 1 such that 3, 3log_y x, 3log_z y, 7log_x z are in A.P. with common difference 1/2. Then r - p - q is equal to
2023
- If dfrac1^3 + 2^3 + 3^3 + … up to n terms1·3 + 2·5 + 3·7 + … up to n terms = 9/5, then the value of n is
2023 · with solution
- Suppose f is a function satisfying f(x+y) = f(x) + f(y) for all x, y ∈ N and f(1) = 1/5. If sum_n=1^m (f(n))/(n(n+1)(n+2)) = 1/12, then m is equal to _______
2023 · with solution
- For the two positive numbers a, b, if a, b and 1/18 are in a geometric progression, while 1/a, 10 and 1/b are in an arithmetic progression, then 16a + 12b is equal to ___________.
2023 · with solution
- Let a_1, a_2, a_3, … be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then a_1 a_9 + a_2 a_4 a_9 + a_5 + a_7 is equal to ________.
2023 · with solution
- The 4^th term of GP is 500 and its common ratio is 1/m, m ∈ N. Let S_n denote the sum of the first n terms of this GP. If S_6 > S_5 + 1 and S_7 < S_6 + 1/2, then the number of possible values of m is
2023 · with solution
- Let A_1, A_2, A_3 be the three A.P. with the same common difference d and having their first terms as A, A+1, A+2, respectively. Let a, b, c be the 7^th, 9^th, 17^th terms of A_1, A_2, A_3, respectively such that…
2023
- Let a, b and c be in GP with common ratio r, where a ≠ 0 and 0 < r ≤ 1/2. If 3a, 7b and 15c are the first three terms of an AP, then the 4th term of this AP is
2019 · with solution
- If a_1, a_2, a_3, …, a_n are in AP and a_1 + a_4 + a_7 + … + a_16 = 114, then a_1 + a_6 + a_11 + a_16 is equal to
2019 · with solution
- The sum of series 1 + (1^3 + 2^3)/(1 + 2) + (1^3 + 2^3 + 3^3)/(1 + 2 + 3) + … + (1^3 + 2^3 + 3^3 + … + 15^3)/(1 + 2 + 3 + … + 15) - 1/2(1 + 2 + 3 + … + 15) is equal to
2019 · with solution
- The sum sum_k=1^20 k · 1/(2^k) is equal to
2019
- If 19th term of a non-zero AP is zero, then its (49th term) : (29th term) is
2019 · with solution
- Let S_n denote the sum of the first n terms of an AP. If S_4 = 16 and S_6 = -48, then S_10 is equal to
2019 · with solution
- The sum of series (3 × 1^3)/(1^2) + (5 × (1^3 + 2^3))/(1^2 + 2^2) + (7 × (1^3 + 2^3 + 3^3))/(1^2 + 2^2 + 3^2) + … upto 10th term, is
2019
- Let S_n = 1 + q + q^2 + ... + q^n and T_n = 1 + ((q+1)/2 ) + ((q+1)/2 )^2 + ... + ((q+1)/2 )^n, where q is a real number and q ≠ 1. If ^101C_1 + ^101C_2 · S_1 + ... + ^101C_101 · S_100 = α T_100, then α is equal to
2019
- The product of three consecutive terms of a GP is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an AP. Then, the sum of the original three terms of the given GP is
2019 · with solution
- The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27/19. Then, the common ratio of this series is
2019
- If the sum and product of the first three terms in an AP are 33 and 1155, respectively, then a value of its 11th term is
2019 · with solution
- Let a_1, a_2, …, a_10 be a GP. If a_3/a_1 = 25, then a_9/a_5 equals
2019 · with solution
- Let S_k = (1+2+3+…+k)/k. If S_1^2 + S_2^2 + … + S_10^2 = 5/12A, then A is equal to
2019
- Let the sum of the first n terms of a non-constant AP a_1, a_2, a_3, … be 50n + (n(n-7))/2A, where A is a constant. If d is the common difference of this AP, then the ordered pair (d, a_50) is equal to
2019 · with solution
- If a, b and c be three distinct real numbers in GP and a + b + c = xb, then x cannot be
2019 · with solution
- The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is
2019 · with solution
- Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression fracx^m y^n(1 + x^2m)(1 + y^2n) is
2019
- The sum of the following series 1 + 6 + (9(1^2 + 2^2 + 3^2))/7 + (12(1^2 + 2^2 + 3^2 + 4^2))/9 + (15(1^2 + 2^2 + ·s + 5^2))/11 + … up to 15 terms is
2019
- Let a_1, a_2, …, a_30 be an AP, S = sum_i=1^30 a_i and T = sum_i=1^15 a_(2i-1). If a_5 = 27 and S - 2T = 75, then a_10 is equal to
2019 · with solution
- If 5, 5r, 5r² are the lengths of the sides of a triangle, then r cannot be equal to
2019 · with solution
- Let a_1, a_2, a_3, …, a_49 be in AP such that sum_k=0^12 a_4k+1 = 416 and a_9 + a_43 = 66. If a_1^2 + a_2^2 + … + a_17^2 = 140 m, then m is equal to
2018
- Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 1^2 + 2 · 2^2 + 3^2 + 2 · 4^2 + 5^2 + 2 · 6^2 + … If B - 2A = 100λ, then λ is equal to
2018
- For any three positive real numbers a, b and c, if 9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c), then
2017
- If the 2nd, 5th and 9th terms of a non-constant AP are in GP, then the common ratio of this GP is
2016
- Let b_i > 1 for i = 1, 2, …, 101. Suppose log_e b_1, log_e b_2, …, log_e b_101 are in AP with the common difference log_e 2. Suppose a_1, a_2, …, a_101 are in AP such that a_1 = b_1 and a_51 = b_51. If t = b_1 + b_2 + ……
2016
- If the sum of the first ten terms of the series (13/5 )^2 + (22/5 )^2 + (31/5 )^2 + 4^2 + (44/5 )^2 + … is 16/5m, then m is equal to
2016
- If m is the AM of two distinct real numbers l and n (l, n > 1) and G_1, G_2 and G_3 are three geometric means between l and n, then G_1^4 + 2G_2^4 + G_3^4 equals
2015
- The sum of first 9 terms of the series (1^3)/1 + (1^3 + 2^3)/(1+3) + (1^3 + 2^3 + 3^3)/(1+3+5) + … is
2015 · with solution
- Three positive numbers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is
2014 · with solution
- If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9, then k is equal to
2014
- The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ... is
2013
- If a_1, a_2, a_3, … are in a harmonic progression with a_1 = 5 and a_20 = 25. Then, the least positive integer n for which a_n < 0, is
2012 · with solution
- If the sum of first n terms of an AP is cn^2, then the sum of squares of these n terms is
2009 · with solution
- Let f(x) = ax^2 + bx + c, a ≠ 0 and Δ = b^2 - 4ac. If α + β, α^2 + β^2 and α^3 + β^3 are in GP, then
2005
- An infinite GP has first term x and sum 5, then x belongs to
2004 · with solution
- Let a, b, c be in an AP and a^2, b^2, c^2 be in GP. If a < b < c and a + b + c = 3/2, then the value of a is
2002 · with solution
- If a_1, a_2, …, a_n are positive real numbers whose product is a fixed number c, then the minimum value of a_1 + a_2 + … + a_n-1 + 2a_n is
2002 · with solution
- If the positive numbers a, b, c, d are in AP. Then, abc, abd, acd, bcd are
2001
- Let α, β be the roots of x^2 - x + p = 0 and γ, δ be the roots of x^2 + 4x + q = 0. If α, β, γ, δ are in GP, then the integer values of p and q respectively are
2001
- If the sum of the first 2n terms of the AP series 2, 5, 8, … is equal to the sum of the first n terms of the AP series 57, 59, 61, …, then n equals
2001 · with solution
- Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 3/4, then
2000 · with solution
- Let a_1, a_2, …, a_10 be in AP and h_1, h_2, equal to …, h_10 be in HP. If a_1 = h_1 = 2 and a_10 = h_10 = 3, then a_4 h_7 is
1999 · with solution
- The harmonic mean of the roots of the equation (5 + √(2))x^2 - (4 + √(5))x + 8 + 2√(5) = 0 is
1999 · with solution
- If x > 1, y > 1, z > 1 are in GP, then 1/(1 + ln x), 1/(1 + ln y), 1/(1 + ln z) are in
1998 · with solution
- The product of n positive numbers is unity, then their sum is
1991
- If a, b and c are distinct positive numbers, then the expression (b+c-a)(c+a-b)(a+b-c)-abc is
1991 · with solution
- If a, b, c, d and p are distinct real numbers such that (a^2 + b^2 + c^2) p^2 - 2(ab + bc + cd) p + (b^2 + c^2 + d^2) ≤ 0, then a, b, c, d
1987 · with solution
- If a, b, c are in GP, then the equations ax^2 + 2bx + c = 0 and dx^2 + 2ex + f = 0 have a common root, if d/a, e/b, f/c are in
1985
- The third term of a geometric progression is 4. The product of the first five terms is
1982 · with solution
- If x_1, x_2, …, x_n are any real numbers and n is any positive integer, then
1982
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