Question

the sum of four consecutive terms which are in an Ap is 32 and the ratio of the product of the first and last term to the product of two middle terms is 7:15 .find the number
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Answer 1

Answer:

Step-by-step explanation:

Let no's are

(a-3d),(a-d),(a+d),(a+3d)

Sum=a-3d+a-d+a+d+a+3d

32=4a

a=8

A/Q (a-3d)(a+3d):(a-d)(a+d) =7:15

(64-9d^2):(64-d^2)=7:15

7(64-d^2)=15(64-9d^2)

=448-7d^2 = 960 - 135d^2

=135d^2- 7d^2=960-448

=128d^2. = 512

d^2 =512/128=4

d=+2 or -2

No. Are, 2,6,10,14 or 14,10,6,2

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