Question

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes (1st and 4th) to the
product of middle (2nd and 3rd) is 5:6.

any other methods to solve it
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please help me fast its very important please​

Answer 1

Answer:

Let 4 parts in AP be a, a +d, a+2d, a+3d

It is given

a+(a+d)+(a+2d)+(a+3d)=56

4a+6d=56

2a+3d=28

{a×(a+3d)}/{(a+d)(a+2d)}

Cross multiply then simplify

a^2+3ad-10d^2=0

Factorise by mid term splitting

(a+5d)(a-2d)=0

a=-5d

a=2d

put this value of a in equation

2a+3d=28

-10d+3d=28

-7d=28

d=-4

a=20

So AP 20, 16,12,8

You will get another set when you take another value of d