1 + 3 + 52 + 7 + 92 + up to 40 terms is equal to
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Previous year questions with solutions · 2025 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 32 | 19 | 1.7 | 6.74% ↑6.4% |
| 2025 | 19 | 12 | 1.6 | 6.33% |
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 :