Question

Four numbers are in G.P. whose sum is $85$ and product is $4096$, the numbers are

• A. 1, 16, 64, 4
• B. 1, 4, 16, 64
• C. 64, 1, 4, 16
• D. 64, 4, 16, 1

Answer 1

The correct option is B 1, 4, 16, 64

Let the numbers be $\frac{a}{r^3},\frac{a}{r},ar,a{r}^3$

Given $\frac{a}{r^3}+\frac{a}{r}+ar+a{r}^3=85$____(1)

$a^4=4096$

$\Rightarrow a=8$

From (1), $\frac{1}{r^3}+\frac{1}{r}+r+r^3=\frac{85}{8}$

$\Rightarrow \frac{1}{r^3}(1+r^2)+r(1+r^2)=\frac{85}{8}$

$\Rightarrow (1+r^2)\left(\frac{1+r^4}{r^3}\right)=\frac{85}{8}$

Clearly, $r=2$ satisfies

$\therefore$ the numbers in GP are $1,4,16,64$