Question

Find n if the sum of the first n terms of the series $\sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+......is435\sqrt{3}$.

Answer 1

$\sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+.......=435\sqrt{3}$

$\Rightarrow \sqrt{3}+\sqrt{25\times 3}+\sqrt{81\times }+\sqrt{169\times 3}.....=435\sqrt{3}$

$\Rightarrow \sqrt{3}+5\sqrt{3}+9\sqrt{3}+13\sqrt{3}.....=435\sqrt{3}$

$\Rightarrow \sqrt{3}[1+5+9+13+.....]=435\sqrt{3}$

$\Rightarrow 1+5+9+13+.....=435$

above series are in A.P

Here a=1, d=4

$\therefore Sn=\frac{n}{2}[2a+(n-1\left)d\right]$

$\therefore \frac{n}{2}[2a+(n-1\left)d\right]=435$

$\frac{n}{2}[2\times 1+(n-1\left)4\right]=435$

$\therefore n[1+2n-2]=435$

$n(2n-1)=435$

$2n^2-n-435=0$

$n=\frac{1\pm \sqrt{1+4\times 2\times 435}}{2.2}=\frac{1\pm 59}{4}$

$\therefore n=15,n=-58/4$
$\therefore n=15$