Question

Divide $20$ into four parts which are in arithmetic progression such that the product of the first and fourth is to the product of the second and third is in the ratio $2:3$ then least value of them is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is A $2$

$a,a+d,a+2d,a+3d\to 20$ in parts of A.P.

Given $\frac{a(a+3d)}{(a+d)(a+2d)}=\frac{2}{3}$

$3a^2+9ad=2[a^2+2ad+ad+2d^2]$

$3a^2+9ad=2a^2+6ad+4d^2$

$a^2+3ad-4d^2=0$

$a^2+4ad-ad-4d^2=0$

$a(a+4d)-d(a+4d)=0$

$(a-d)(a+4d)=0$

$\Rightarrow a=d$ (or) $a=-4d$

Let $a=d$ the sum $a+2a+3a+4a=20$

$a=2$ least value.