Question

the sum of four consecutive numbers in an ap is 32 and the ratio of the product of the first and the last term to the product of two middle term is 715 find the numbers.​

Answer 1

Hey there!

Here is your answer

Let the AP with 4 terms be:

a-3d, a-d, a+d, a+3d------------(1)

Given, S4=32;

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

4a=32

=a=32/4= 8

Product of first and last term/product of the 2 middle terms= 7/15

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

a^2-9d^2/a^2-d^2)=7/15

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

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

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

=8a^2=128d^2

(a=8)

==> 8(8)^2/128=d^2

=512/128=d^2

=d^2=4

=d= +/-2

Substitute the values of a and d in (1)

And there by AP=

2,6,10,14/ 14,10,6,2

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Hope it helps you!!

Happy learning ✨

Answer 2

plz refer to this attachment