Question

If the sum of the first n terms of the series $\sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+....is435\sqrt{3}$, then $n$ equals :

• A. 18
• B. 15
• C. 13
• D. 29

Answer 1

Answer: (B)

$\sqrt{3}\left[1+\sqrt{25}+\sqrt{81}+\sqrt{69}+......\right]=435\sqrt{3}$
$\sqrt{3}\left[1+5+9+13+.....T_n\right]=435\sqrt{3}$
$=\sqrt{3}\times \frac{n}{2}\left[2+\left(n-1\right)4\right]=435\sqrt{3}$
$2n+4n^2-4n=870<br/>=4n^2-2n-870=0$
$=2n^2-n-435=0$
$n=\frac{1\pm \sqrt{1+4\times 2\times 435}}{4}$
$=\frac{1\pm 59}{4}$
$=\frac{1+59}{4}=4;\frac{1-59}{4}$