Question

$\sqrt{x}$, $\sqrt{2x}$, $2\sqrt{x}$, $2\sqrt{2x}$ $\dots \dots$ are in geometric progression.

If the sum of first 10 terms is $31(\sqrt{6}+\sqrt{3})$, then find the 10th term.

Answer 1

Given

$\sqrt{x},\sqrt{2}\sqrt{x},2\sqrt{x}.2\sqrt{2}x\to G.P$

Common ratio$=\frac{\sqrt{2}\sqrt{x}}{\sqrt{x}}$

$=\sqrt{2}$

Given,

sum of first $10$ terms$=31(\sqrt{6}+\sqrt{3})$

$S_n=\frac{a_1(1-r^n)}{1-r},r\ne 1$

$S_{10}=\frac{\sqrt{x}(1-(\sqrt{2}{)}^{10})}{1-\sqrt{2}}=31(\sqrt{6}+\sqrt{3})$

$\sqrt{x}\frac{(1-32)}{1-\sqrt{2}}=31(\sqrt{6}+\sqrt{3})$

$\sqrt{x}\times \frac{31}{\sqrt{2}-1}=31(\sqrt{6}+\sqrt{3})$

$\sqrt{x}=\frac{\sqrt{6}+\sqrt{3}}{\sqrt{2}-1}$

On rationalizing

$\sqrt{x}=\frac{(\sqrt{6}+\sqrt{3})(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}$

$\sqrt{x}=\sqrt{6}\sqrt{2}+\sqrt{6}+\sqrt{3}\sqrt{2}+\sqrt{3}$

$T_n=a_1{r}^{n-1}$

$T_{10}=(\sqrt{6}\sqrt{2}+\sqrt{6}+\sqrt{3}\sqrt{2}+\sqrt{3})(\sqrt{2}{)}^9$

$T_{10}=(\sqrt{12}+\sqrt{6}+\sqrt{6}+\sqrt{3})\frac{32}{\sqrt{2}}$

$T_{10}=(\sqrt{6}\sqrt{2}+2\sqrt{6}+\sqrt{3})\frac{32}{\sqrt{2}}$

$T_{10}=(2\sqrt{3}+2\sqrt{6}+\sqrt{3})\times \frac{32}{\sqrt{2}}$

$T_{10}=(3\sqrt{3}+2\sqrt{2}\sqrt{3})\times \frac{32}{\sqrt{2}}$

$T_{10}=16\sqrt{2}(3\sqrt{3}+2\sqrt{6})$.