Sequences and Series · Question 40

The sum 1 + 3 + 11 + 25 + 45 + 71 + upto 20 terms, is equal to

1. 7240
2. 8124
3. 7130
4. 6982

The sum 1 + 1 + 32! + 1 + 3 + 53! + 1 + 3 + 5 + 74! + up to terms, is equal to

1. 3e
2. 2e
3. 4e
4. 6e

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 :

1. 18
2. 17
3. 11
4. 10

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 :

1. 6
2. 4
3. 8
4. 10

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 :

1. 108
2. 90
3. 122
4. 84

The value of n ( k=1n k3 + 6k2 + 11k + 5(k+3)! ) is :

1. 5/3
2. 2
3. 4/3
4. 7/3

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

1. 3a100 + 100
2. 3a100 - 100
3. 3a99 - 100
4. 3a99 + 100

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

1. 98
2. 126
3. 112
4. 142

For positive integers n, if 4an = (n2 + 5n + 6 ) and Sn = k=1n ( 1ak ), then the value of 507S2025 is :

1. 540
2. 675
3. 1350
4. 135