Question

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first
and last terms to the product of two middle terms is 7:15. Find the number.​

Answer 1

Answer:

Let the 4 consecutive numbers in AP be

(a−3d),(a−d),(a+d),(a+3d)

According to the question,

a−3d+a−d+a+d+a+3d=32

4a=32

a=8

Now,

(a−d)(a+d)

(a−3d)(a+3d)

=

15

7

15(a

2

−9d

2

)=7(a

2

−d

2

)

15a

2

−135d

2

=7a

2

−7d

2

8a

2

=128d

2

Putting the value of a, we get,

d

2

=4

d=±2

So, the four consecutive numbers are 2,6,10,14 or 14,10,6,2.

Step-by-step explanation:

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