The sum of four consecutive terms which are in an arithmetic progression is $32$ and the ratio of the products of the first and last terms to the product of the middle terms is $7:15$. Find the AP.
I know it's possible to solve this by taking the terms as
$$a-3d,a-d,a+d,a+3d$$
And even found an answer which explained that the way you represent terms doesn't matter but I am still not able to solve this question with terms as other variables such as
$1)$ $a-2d,a-d,a+d,a+2d$
$2)$ $a, a+d, a+2d, a+3d$
Can anyone solve this with the terms in $(1)$ and $(2)$ ?
The AP is 2,6,10,14 but how will you find it with (1) or (2)?
If you choose the second case,
the sum will be
$$4a+6d=32$$ or $$3d+2a=16$$
But $$3\times 16 - 2\times 16=16$$
thus $$3(16-d) = 2(a+16)$$ and since $2$ and $3$ are relatively prime,
$$d =16 - 2k \text{ and } a= 3k-16$$
the second equation is
$$\frac{a(a+3d)}{(a+d)(a+2d)}=\frac{7}{15}$$
gives $$\frac{(3k-16)(32-3k)}{k(16-k)}=\frac{7}{15}$$
and $k=6$, $a=2$ , $d=4$. The four terms are $$\boxed{2, 6, 10, 14}$$