Question

Divide $40$ into four parts, which are in $A.P$. such that ratio of product of extremes is to the product of means is $2:3$.

Answer 1

Let the four parts of $40$ are

$a-3d,a-d,a+d,a+3d$

then , $a-3d+a-d+a+d+a+3d=40$

$\Rightarrow 4a=40$

$\Rightarrow a=10$

Now ,

$\frac{(a-3d)\left(a+3d\right)}{\left(a-d\right)\left(a+d\right)}=\frac{2}{3}$

$\Rightarrow \frac{a^2-9d^2}{a^2-d^2}=\frac{2}{3}$

$\Rightarrow \frac{\left({10}^2\right)-9d^2}{{\left(10\right)}^2-d^2}=\frac{2}{3}$

$\Rightarrow \frac{100-9d^2}{100-d^2}=\frac{2}{3}$

$\Rightarrow 300-27d^2=200-2d^2$

$\Rightarrow 100=25d^2$

$\Rightarrow d^2=4$

$\Rightarrow d=\pm 2$

If $d=2,a=10$

then four parts of $40$ are

$10-3\times 2,10-2,10+2,10+3\times 2$

$=4,8,12,16$

If $d=-2,a=10$ then , four parts are

$10-3\times -2,10-\left(-2\right),10-2,10+3(-2)$

$=16,12,8,4$

hence four parts of $40$ are $4,8,12,and16$