Question

The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Answer 1

It is given that the first term of an A.P. is 11 where the sum of first four terms is 56 and sum of last four terms is 112.

Let the A.P. be a,a+d,a+2d,a+3d,…,a+( n−1 )d.

The sum of first four terms is,

a+( a+d )( a+2d )+( a+3d )=4a+6d

The sum of last four terms is,

[ a+( n−4 )d ]+[ a+( n−3 )d ]+[ a+( n−2 )d ]+[ a+( n−1 )d ]=4a+( 4n−10 )d

So, the sum of first four terms becomes,

4a+6d=56 4( 11 )+6d=56 6d=56−44 d=2

So, the sum of last four terms becomes,

4a+( 4n−10 )d=112 4( 11 )+( 4n−10 )2=112 4n−10=34 n=11

Therefore, the number of terms of the A.P.is 11.