Question

VI. Solve T.
38. The sum of four consetive terms of an AP is 32 The ratio of the product of first
and last term and product of middle two terms in 7:5 Find the terms of an AP.​

Answer 1

Let 4 consecutive term of AP be a-3d,a-d,a+d,a+3d.

Then,

a+3d+a+d+a-d+a-3d=32

4a=32

a=8

Now,

product of first and last term=(a+3d)(a-3d)

= a²-9d²

product of middle two term=(a+d)(a-d)

=a²-d²

Now,

a²-9d²/a²-d²=7/5

5(a²-9d²)=7(a²-d²)

5a²-45d²=7a²-7d²

-2a²=38d²

-2(8)²=38d²

-128=38d²

d=under root -128/38