Question

Find the seventh term of the G.P.: $\sqrt{3}+1,1,\frac{\sqrt{3}-1}{2}$....................

Answer 1

$\sqrt{3}+1,1,\frac{\sqrt{3}-1}{2}$

$\Rightarrow r=\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{\sqrt{3}-1}{2}$

Seventh term $=a{r}^6$

$=\sqrt{3}+1\times {\left(\frac{\sqrt{3}-1}{2}\right)}^6$

$=(\sqrt{3}+1)(\sqrt{3}-1)\frac{{(\sqrt{3}-1)}^5}{2^6}$

$=\frac{2{(\sqrt{3}-1)}^5}{2^6}$

$=\frac{{(\sqrt{3}-1)}^5}{2^5}$

$=\frac{{(3-2\sqrt{3}+1)}^2(\sqrt{3}-1)}{2^5}$

$=\frac{{(4-2\sqrt{3})}^2(\sqrt{3}-1)}{2^5}$

$=\frac{2^2\left(2\sqrt{3}\right)(\sqrt{3}-1)}{2^5}$

$=\frac{(4-4\sqrt{3}+3)(\sqrt{3}-1)}{2^3}$

$=\frac{(7-4\sqrt{3})(\sqrt{3}-1)}{2^3}$

$=\frac{7\sqrt{3}-7-12+4\sqrt{3}}{2^3}$

$=\frac{11\sqrt{3}-19}{2^3}$

Hence, the answer is $\frac{11\sqrt{3}-19}{2^3}.$