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 Multiple Choice QuestionsMultiple Choice Questions

231.

If S is the sum of the first 10 terms of the series tan-113 + tan-117  + tan-1113 + tan-113 + tan-1121 + ... , then tanS = ?

  • - 65

  • 511

  • 56

  • 1011


232.

If 210 + 29 . 31 +28 . 32 +... + 2 . 39 +310 = 310= S - 211 Then S = ?

  • 311

  • 2 . 311

  • 3112 + 210

  • 311 - 211


233.

If the sum of the second,third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

  • 213350 - 1

  • 126349 - 1

  • 113350 - 1

  • 126350 - 1


234.

If the sum of the first 20 terms of the series loglog712x + log713x + log712x + ... is 460,

then x = ?

  • e2

  • 74621

  • 72

  • 712


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 Multiple Choice QuestionsShort Answer Type

235.

The coefficient of x4 in the expansion of (1 + x + x2 + x3)6 in powers of x, is .


 Multiple Choice QuestionsMultiple Choice Questions

236.

Out of 11 consecutive natural number if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference is :

  • 1099

  • 15101

  • 533

  • 5101


237.

Let a , b, c, d and ay non zero distinct real numbers such that (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :

  • a, b, c, d are in A.P.

  • a, c, p are in G.P.

  • a, b, c, d are in G.P.

  • a, c, p, are in A.P.


238.

The 100th term of the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ... , is

  • 12

  • 13

  • 14

  • 15


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239.

Let Sn denotes the sum of first n terms of an AP.If S4 = - 34, S5 = - 60 and S6 = - 93, then the common difference and the first term of the AP are respectively

  • - 7, 2

  • 7, - 4

  • 7, - 2

  • - 7, - 2


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240.

An AP has the property that the sum of first ten terms is half the sum of next ten terms. If the second term is 13, then the common difference is

  • 3

  • 2

  • 5

  • 4


B.

2

Let a and d be the first term and common difference, respectively.

Given, T2 = 13  a + d = 13        ...(i)

According to the question,

a1 +a2  ... + a10 = 12a11 + a12 + a20 2a1 +a2 + ... + a10 = a11 + a12 + a20 3a1 +a2 + ... + a10 = a1 +a2 + ... + a20add  a1 + a2 + ... + a10 on both sides 3 × 1022a + 9d = 2022a + 19d 6a + 27d = 4a + 38d 2a = 11d                     ...iiFrom Eqs. (i) and (ii), we get11d2 + d = 13  d = 2


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