Sum of 4 consecutive terms of an A.P is 32. The ratio of product of 1s

1.	Sum of 4 consecutive terms of an A.P is 32. The ratio of product of 1st and last term andproduct of 2nd and 3rd term is 7:15. Find the terms.
Answer» Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d. a-3d+a-d+a+d+a+3d=32 4a=32 a=8 1 (a-3d)(a=3d) /(a-d)(a+d)=7/15 15(a2-9d2) = 7(a2-d2) 15a2-135d2 =7a2-7d2 8a2-128d2= 0 d2= 4, ±2 Therefore d=4 or d=±2 1m So, when a=8 and d=2, the numbers are 2,6,10,14. When a=8,d=-2 the numbers are 14,10,6,2

Sum of 4 consecutive terms of an A.P is 32. The ratio of product of 1st and last term andproduct of 2nd and 3rd term is 7:15. Find the terms.

Answer»

Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d.

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

4a=32

a=8 1

(a-3d)(a=3d) /(a-d)(a+d)=7/15

15(a2-9d2) = 7(a2-d2)

15a2-135d2 =7a2-7d2

8a2-128d2= 0

d2= 4, ±2

Therefore d=4 or d=±2 1m

So, when a=8 and d=2, the numbers are 2,6,10,14.

When a=8,d=-2 the numbers are 14,10,6,2