Q37 of 84 Page 5

Find the sum of first n terms of an A.P. whose nth term is 5n – 1. Hence find the sum of first 20 terms. (CBSE 2011)

n^th term = 5n - 1

Put n=1, we get

1^st term = 5-1 = 4

Hence sum of n terms = n(First term + nth term)/2

= n(4+5n - 1)/2

= n(5n +3)/2

Sum of 20 terms becomes, put n = 20

S₂₀ = [20(5×20 +3)]/2

=10(103) = 1030

Find an A.P. whose fourth term is 9 and the sum of its sixth term and the thirteenth term is 40. (CBSE 2011)

Find the value of the middle term of the following A.P.: -6, -2, 2, …, 58. (CBSE 2011)

Determine the A.P. whose fourth term is 18 and the difference of the ninth term from the fifteenth term is 30. (CBSE 2011)

Find the sum of all multiples of 7 lying between 500 and 900. ? (CBSE 2012)

The 17th term of AP is 5 more than twice its 8th term. If the 11th term of the AP is 43, find its nth term? (CBSE 2012)