Sequences and Series · Question 45

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

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 .