Question

the sum of four consecutive numbers in an ap is 32 and ratio of the product of first and last term to the product of two middle term is 7:15. find the numbers?

Answer 1

let the four consecutive numbers be

a-3d,a-d,a+d,a+3d

then according to ques.

a-3d+a-d+a+d+a+3d= 32
so 4a = 32
a= 8

and ratio of product of extreme and middle is 7/15

i.e (a-3d)(a+3d)/(a-d)(a+d) = 7/15


but a = 8
so (8-3d)(8+3d)/(8-d)(8+d) = 7/15

64 - (3d^2)/64-d^2 = 7/15

64-9d^2/64-d^2= 7/15
(64-9d^2)*15 = (64-d^2)*7
960-135d^2 = 448 - 7d^2

960-448 = 135d^2- 7d^2
512= 128d^ 2
d= 4

so numbers are 2,6,10,14


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Answer 2

and after this by the same formula you can take out Sn/Sl by putting the ratio.
Then by simultaneously you can find out the answer