Question

In an arithmetic progression the 11th term is three times the 2nd term and the twenty-fifth term is 135.

i)Find the sum of the first 100 terms.

(ii) Find the formula for the nth term of this sequence. ​

Answer 1

Explanation:

let a be the first term

x be the increment

2nd term will be a+x

3rd term will be a+2x

so 25 th term = a+24x= 135

11th term

a+10x = 3(a+x)

a = 7/2 x

solve this in 25th term

x = 10

a= 35

sum of 100 terms

s= n/2(35+1025)

1025 is the 100th term and n = 100

s = 53000

nth term

35+10(n-1)