Let =3+4+8+9+13+14+ upto 40 terms. If ( ) 1020 is a root of the equation x2+x-2=0, (0, 2 ), then 2 +3 2 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% |
Let =3+4+8+9+13+14+ upto 40 terms. If ( ) 1020 is a root of the equation x2+x-2=0, (0, 2 ), then 2 +3 2 is equal to :
The value of 13-23+33- +153 is :
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The sum 1+ 12 (12+22 )+ 13 (12+22+32 )+ upto 10 terms is equal to :
Let the sum of the first n terms of an A.P. be 3n2+5n. Then the sum of squares of the first 10 terms of the A.P. is:
n=110 ( 528n(n+1)(n+2) ) is equal to:
If the sum of the first 10 terms of the series 11+14 4+ 21+24 4 + 31+34 4+ 41+44 4+ is mn, (m, n)=1, then m+n is equal to :
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The first term of an A.P. of 30 non-negative terms is 103. If the sum of this A.P. is the cube of its last term, then its common difference is:
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Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in A B, which are divisible by 3, is :
Let a1, a2, a3, be an A.P. and g1=a1, g2, g3, be an increasing G.P. If a1=a2+g2=1 and a3+g3=4, then a10+g5 is equal to:
The sum 131+ 13+231+3+ 13+23+331+3+5+ up to 8 terms, is :
The value of k=1 (-1)k+1 ( k(k+1)k! ) is
The common difference of the A.P.: a1, a2, , am is 13 more than the common difference of the A.P.: b1, b2, , bn. If b31 = -277, b43 = -385 and a78 = 327, then a1 is equal to
6326 + 10 1325 + 10 2324 + 10 22323 + + 10 2243 is equal to :
Let 729, 81, 9, 1, be a sequence and Pn denote the product of the first n terms of this sequence. If 2 n=140(Pn) 1n = 3 -13 and ( , ) = 1, then + is equal to
Consider an A.P.: a1, a2, , an; a1 > 0. If a2 - a1 = -34, an = 14a1, and i=1n ai = 5252, then i=117 ai is equal to
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( 13 + 47 ) + ( 132 + 13 47 + 4272 ) + ( 133 + 132 47 + 13 4272 + 4373 ) + up to infinite terms, is equal to
Let k=1n ak = n2 + n. If a10 = 59 and a6 = 7a1, then + is equal to :
If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
Let a1, a2, a3, be a G.P. of increasing positive terms such that a2 a3 a4 = 64 and a1 + a3 + a5 = 8137. Then a3 + a5 + a7 is equal to :
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Let a1, a22, a322, , a1029 be a G.P. of common ratio 1 2. If a1 + a2 + + a10 = 62, then a1 is equal to:
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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 :