Question

The sum of four consecutive numbers in an AP is 32 . If the ratio of product of first and last term to the product of middle terms is 7:15 . Find the numbers .

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Let the no.s in AP be a-3d,a-d,a+d & a+3d

Acc to question,

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

4a=32

a=8

Also,

(a-3d)(a+3d). 7

----------------- =. - - -

(a-d) (a+d) 15

a^2-9d^2. 7

--------------- = - - - - -

a^2-d^2. 15

15a^2-135d^2=7a^2-7d^2

8a^2=128d^2

8(64)=128d^2

d^2=512/128

d^2=4

d=±2

Therefore

The no.s are. Or. The no.s are

2,6,10,14. 14,10,6,2

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