1 + 3 + 52 + 7 + 92 + up to 40 terms is equal to
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Previous year questions with solutions · 2024 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 32 | 19 | 1.7 | 6.74% ▲ 14.3% vs last year |
| 2025 | 28 | 19 | 1.5 | 5.89% ▲ 1.1% vs last year |
| 2024 | 35 | 20 | 1.8 | 5.83% |
1 + 3 + 52 + 7 + 92 + up to 40 terms is equal to
If the sum of the first 20 terms of the series 4 14 + 3 12 + 14 + 4 24 + 3 22 + 24 + 4 34 + 3 32 + 34 + 4 44 + 3 42 + 44 + is mn, where m and n are coprime, then m + n is equal to :
Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of A.P.'s in A and B respectively such that D=d+3, d>0. If p+qp-q= 195, then p-q is equal to
Let a1, a2, a3, be a G.P. of increasing positive numbers. If a3 a5 = 729 and a2 + a4 = 1114, then 24(a1 + a2 + a3) is equal to
The sum 1 + 3 + 11 + 25 + 45 + 71 + upto 20 terms, is equal to
The sum 1 + 1 + 32! + 1 + 3 + 53! + 1 + 3 + 5 + 74! + up to terms, is equal to
Let a1, a2, a3, be in an A.P. such that k=112 a2k-1 = - 725a1, a1 0. If k=1n ak = 0, then n is :
The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by 212. Then the number of terms which are integers in the A.P. is :
Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :
The value of n ( k=1n k3 + 6k2 + 11k + 5(k+3)! ) is :
Let an be a sequence such that a0 = 0, a1 = 12 and 2an+2 = 5an+1 - 3an, n = 0, 1, 2, 3, . Then k=1100 ak is equal to
Let Tr be the rth term of an A.P. If for some m, Tm = 125, T25 = 120, and 20 r=125Tr = 13, then 5m r=m2mTr is equal to
For positive integers n, if 4an = (n2 + 5n + 6 ) and Sn = k=1n ( 1ak ), then the value of 507S2025 is :
Let Sn= 12+ 16+ 112+ 120+ upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is 2026 S2025, then the absolute difference between 20th and 15th terms of the A.P. is
In an arithmetic progression, if S40=1030 and S12=57, then S30-S10 is equal to :
If 7=5+ 17(5+ )+ 172(5+2 )+ 173(5+3 )+ , then the value of is :
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to
Let a1,a2,a3, be a G.P. of increasing positive terms. If a1a5=28 and a2+a4=29, then a6 is equal to:
If r=1nTr= (2n-1)(2n+1)(2n+3)(2n+5)64, then n r=1n ( 1Tr ) is equal to :
Suppose that the number of terms in an A.P. is 2k, k . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to:
If the sum of the series 11 (1 + d) + 1(1 + d)(1 + 2d) + + 1(1 + 9d)(1 + 10d) is equal to 5, then 50d is equal to :
Let a, ar, ar2, be an infinite G.P. If n=0 arn = 57 and n=0 a3r3n = 9747, then a + 18r is equal to
In an increasing geometric progression of positive terms, the sum of the second and sixth terms is 703 and the product of the third and fifth terms is 49. Then the sum of the 4th, 6th and 8th terms is equal to:
n (12 - 1)(n - 1) + (22 - 2)(n - 2) + + ((n - 1)2 - (n - 1)) 1(13 + 23 + + n3) - (12 + 22 + + n2) is equal to :
Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is the sum of areas of all the triangles formed in this process, then :
A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of m is equal to:
If 1 1 + 2 + 1 2 + 3 + + 1 99 + 100 = m and 11 2 + 12 3 + + 199 100 = n, then the point (m, n) lies on the line
For x 0, the least value of K, for which 41 + x + 41 - x, K2, 16x + 16-x are three consecutive terms of an A.P., is equal to :
Let the first three terms 2, p and q, with q 2, of a G.P. be respectively the 7th, 8th and 13th terms of an A.P. If the 5th term of the G.P. is the nth term of the A.P., then n is equal to:
The value of 1 22 + 2 32 + + 100 (101)212 2 + 22 3 + + 1002 101 is