Question

The sum of 4 consecutive terms are in a.p whose sum is 32 and the ratio of the product of the first & last term to the product of middle 2 terms is 7:15 find the numbers?​

Answer 1

Answer:

Let the four consecutive numbers in AP be (a−3d),(a−d),(a+d) and (a+3d)

So, according to the question.

a−3d+a−d+a+d+a+3d=32

4a=32

a=32/4

a=8......(1)

Now, (a−3d)(a+3d)/(a−d)(a+d)=7/15

15(a²−9d²)=7(a²−d²)

15a²−135d²=7a²−7d²

15a²−7a²=135d²−7d² 

8a²=128d²

Putting the value of a=8 in above we get.

8(8)²=128d²

128d²=512

d²=512/128

d²=4

d=2

So, the four consecutive numbers are 

8−(3×2)

8−6=2

8−2=6

8+2=10

8+(3×2)

8+6=14

Four consecutive numbers are 2, 6, 10 and 14.

Answer 2

Step-by-step explanation:

Let the nos. be a-3d, a-d, a+d, a+3d

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

4a=32

a=8

Now,

(8-3d)(8+3d) = 7 [a=8]

(a-d)(a+d) 15

64-9d² = 7

64-d² 15

d=+ or - 2

Hence the numbers are: 2,6,10,14