Question

find the sum of first 22 terms of an ap in which 4th term is 15 and 8th term is one more than twice the 4th term

Answer 1

let the first term =a , and common difference = d of AP
1)
4th term =15
a+3d =15 ---(1)
8th term =1 more than twice the 4th term
a+7d= 2(a+3d)+1
a+7d =2a +6d+1
-1 = 2a +6d-a-7d
-1 =a-d
a-d = -1---(2)
subtract (2) from (1) we get
d= 4
substitute d=4 in (2)
a-d = -1
a-4 = -1
a = -1+4
a = 3
2) a=3 , d=4, n= 22
sn = n/2[2a+(n-1)d]
s22= 22/2[2*3+(22-1)4]

= 11[6+21*4]
=11[6+84]
=11*90
= 990