Question

There are two sets A and B of three numbers in A.P. whose sum is $15$ and where D and d are the common differences such that $D-d=1$. Find the numbers if $\frac{P}{p}=\frac{7}{8}$ where P and p are the products of the numbers in the two sets.

Answer 1

Here $4a=28$
$\therefore a=7$
Also $\frac{(a-3d)(a+d)}{(a-d)(a+3d)}=\frac{8}{15}$
or $15[a^2-3d^2-2ad]=8[a^2-3d^2+2ad]$
or $7[a^2-3d^2]=46ad$
or $7(49-3d^2)=46\times 7$.d
or $49-3d^2=46d$ or $3d^2=46d-49=0$
$(d-1)(3d+49)=0$
$\therefore d=1$.
$\therefore$ Required numbers are $4,6,8,10$.