Question

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

Answer 1

Let four terms in $A.Pa-3d,a-d,a+d,a+3d$

Gicen: $\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{5}{6}$...(1)

$a-3d+a-d+a+d+a+3d=56$

$4a=56$

$a=14$

Put $a=14$ in eq. (1)

$\frac{(14-3d)(14+3d)}{(14-d)(14+d)}=\frac{5}{6}$

$\frac{196-9d^2}{196-d^2}=\frac{5}{6}$

$6(196-9d^2)=5(196-d^2)$

$1176-54d^2=980-5d^2$

$1176-980=-5d^2+54d^2$

$196=49d^2$

$\frac{(14{)}^2}{(7{)}^2}=d^2$

$d=\pm 2$

A.P when $a=14$, $d=2$

$8,12,16,20$

A.P when $a=14$, $d=-2$

$20,16,12,8$