Sequences and Series · Question 6

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:

1. 10220
2. 12860
3. 15220
4. 19780

n=110 ( 528n(n+1)(n+2) ) is equal to:

1. 65
2. 130
3. 220
4. 440

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 :

1. 256
2. 264
3. 276
4. 284

Let A1, A2, A3, , A39 be 39 arithmetic means between the numbers 59 and 159. Then the mean of A25, A28, A31 and A36 is equal to :

1. 129
2. 136
3. 131.50
4. 134

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:

1. 587
2. 2583
3. 1529
4. 529

Let = 14+ 18+ 116+ and = 13+ 19+ 127+ . Then the value of (0.2) 5( )+(0.04) 5( ) is equal to :

1. 4
2. 5
3. 8
4. 25

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 :

1. 4
2. 5
3. 6
4. 7

If k=1n ak=6n3, then k=16 ( ak+1-ak36 )2 is equal to .

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:

1. 81
2. 76
3. 62
4. 55

The sum 131+ 13+231+3+ 13+23+331+3+5+ up to 8 terms, is :

1. 70
2. 71
3. 72
4. 73