Divide 56 in four parts in AP such that the ratio of product of their extremes 1st and 4th to product of means 2nd and 3rd is 5:6

Question

the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d

Answer 1

You can call the four numbers $a$, $a+d$, $a+2d$, $a+3d$ if you like. This won't affect the final answer although it might make the intermediate algebra more complex.

So we have

$6a(a+3d) = 5(a+d)(a+2d)$

$\Rightarrow 6a^2 + 18ad = 5a^2 + 15ad + 10d^2$

$\Rightarrow a^2 + 3ad - 10d^2 = 0$

$\Rightarrow (a+5d)(a-2d) =0$

So $a=-5d$ or $a=2d$. We also know that the sum of the four terms is 56, so

$4a+6d=56$

From this you can find two alternative pairs of values for $a$ and $d$. In fact, these represent the same four terms, but one sequence is the reverse of the other. This makes sense given the symmetry of the original problem.