Question

The sum of the first n terms of the arithmetical progression 3, 5.5, 8.!.... is equal to the 2nth term
of the arithmetical progression 16.5, 28.5, 40.5.
Calculate the value of n.​

Answer 1

Answer:

The sum of first 2n terms of 2,5,8,... is

=22n[2(2)+(2n−1)(3)]

=n(6n+1)

The sum of n terms of 57,59,61,... is

=2n[2(57)+(n−1)(2)]

=n(56+n)

Also, n(6n+1)=n(56+n)

⇒6n+1=56+n

⇒5n=55

⇒n=11