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 :
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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% |
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
If ( 1 + 1 + 1 + 2 + + 1 + 1012 ) - ( 12 1 + 14 3 + 16 5 + + 12024 2023 ) = 12024, then is equal to .
Let the positive integers be written in the form : [figure] If the kth row contains exactly k numbers for every natural number k, then the row in which the number 5310 will be, is .
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:
An arithmetic progression is written in the following way [figure] The sum of all the terms of the 10th row is .
Let the first term of a series be T1 = 6 and its rth term Tr = 3Tr-1 + 6r, r = 2, 3, , n. If the sum of the first n terms of this series is 15 (n2 - 12n + 39 ) (4 6n - 5 3n + 1 ), then n 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 S(x) = (1 + x) + 2(1 + x)2 + 3(1 + x)3 + + 60(1 + x)60, x 0, and (60)2 S(60) = a(b)b + b, where a, b N, then (a + b) 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
Let a1, a2, a3, be in an arithmetic progression of positive terms. Let Ak = a12 - a22 + a32 - a42 + + a2k-12 - a2k2. If A3 = -153, A5 = -435 and a12 + a22 + a32 = 66, then a17 - A7 is equal to .
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 :
If 1 + 3 - 22 3 + 5 - 2 618 + 9 3 - 11 236 3 + 49 - 20 6180 + upto = 2 + ( ba + 1 ) e ( ab ), where a and b are integers with (a, b) = 1, then 11a + 18b 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:
Let a=1+ 2C23!+ 3C24!+ 4C25!+ , b=1+ 1C0+1C11!+ 2C0+2C1+2C22! + 3C0+3C1+3C2+3C33!+ . Then 2ba2 is equal to .
The value of 1 22 + 2 32 + + 100 (101)212 2 + 22 3 + + 1002 101 is
Let three real numbers a, b, c be in arithmetic progression and a + 1, b, c + 3 be in geometric progression. If a > 10 and the arithmetic mean of a, b and c is 8, then the cube of the geometric mean of a, b and c is
Let 3, a, b, c be in A.P. and 3, a-1, b+1, c+9 be in G.P. Then, the arithmetic mean of a, b and c is :
Let 3, 7, 11, 15, , 403 and 2, 5, 8, 11, , 404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to .
Let Sn denote the sum of the first n terms of an arithmetic progression. If S10=390 and the ratio of the tenth and the fifth terms is 15:7, then S15-S5 is equal to :
If three successive terms of a G.P. with common ratio r(r>1) are the lengths of the sides of a triangle and [r] denotes the greatest integer less than or equal to r, then 3[r]+[-r] is equal to .
The sum of the series 11-3 12+14+ 21-3 22+24+ 31-3 32+34+ up to 10-terms is
Let 2nd, 8th and 44th terms of a non-constant A. P. be respectively the 1st, 2nd and 3rd terms of a G. P. If the first term of the A. P. is 1, then the sum of its first 20 terms is equal to -
Let Sn denote the sum of first n terms of an arithmetic progression. If S20=790 and S10=145, then S15-S5 is :
Let =12+42+82+132+192+262+ upto 10 terms and = n=110 n4. If 4 - =55k+40, then k is equal to .
Let a and b be two distinct positive real numbers. Let 11th term of a GP, whose first term is a and third term is b, is equal to pth term of another GP, whose first term is a and fifth term is b. Then p is equal to
Let Sn be the sum to n-terms of an arithmetic progression 3,7,11, . If 40< ( 6n(n+1) k=1n Sk )<42, then n equals .
If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to