Question

Divide 32 into four parts which are in AP such that the product of extremes is to the product of means as 7 :15.

Answer 1

Let the required parts be $(a-3d),(a-d),(a+d)$ and $(a+3d)...\left(i\right)$.

Then, $(a-3d)+(a-d)+(a+d)+(a+3d)=32$

$\Rightarrow$ 4a = 32 $\Rightarrow$ $a=8$.

So, the required parts are$(8-3d),(8-d),(8+d)$ and $(8+3d)$

But product of extremes is to the product of means as $7:15$.

$\therefore \frac{(8-3d)(8+3d)}{(8-d)(8+d)}=\frac{7}{15}$

$\Rightarrow \frac{(64-9d^2)}{(64-d^2)}=\frac{7}{15}$

$\Rightarrow 15(64-9d^2)=7(64-d^2)$

$\Rightarrow 128d^2=512\Rightarrow d^2=\frac{512}{128}=4\Rightarrow d=\pm 2$

Thus, (a = 8 and d = 2) or (a = 8 and d = -2).

Hence, the required parts are $2,6,10,14$ or $14,10,6,2$. [From (i)]