The positive integer n, for which the solutions of the equation x(x + 2) + (x + 2)(x + 4) + + (x + 2n - 2)(x + 2n) = 8n3 are two consecutive even integers, is :
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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% |
The positive integer n, for which the solutions of the equation x(x + 2) + (x + 2)(x + 4) + + (x + 2n - 2)(x + 2n) = 8n3 are two consecutive even integers, is :
Let a1, a22, a322, , a1029 be a G.P. of common ratio 1 2. If a1 + a2 + + a10 = 62, then a1 is equal to:
If 114 + 124 + 134 + = 490, 114 + 134 + 154 + = , 124 + 144 + 164 + = , then is equal to :
Let x1, x2, x3, x4 be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from x1, x2, x3, x4, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1 x2 x3 x4) is:
Let an be the nth term of an A.P. If Sn = a1 + a2 + a3 + + an = 700, a6 = 7 and S7 = 7, then an is equal to :
If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :
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 :
If the sum of the first 10 terms of the series 4 11 + 4 14 + 4 21 + 4 24 + 4 31 + 4 34 + is mn, where (m, n) = 1, then m + n is equal to .
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 :
The interior angles of a polygon with n sides are in an A.P. with common difference 6 . If the largest interior angle of the polygon is 219 , then n is equal to .