Question

If the sum of the first n terms of the series $\sqrt{3} + \sqrt{75} + \sqrt{243 }+ \sqrt{507} + .... is \, 435 \sqrt{3}$, then $n$ equals :

• A. 18
• B. 15
• C. 13
• D. 29

Answer 1

Answer: (B)

$\sqrt{3}\left[1+\sqrt{25}+\sqrt{81}+\sqrt{69}+......\right] = 435\sqrt{3}$
$\sqrt{3}\left[ 1 + 5 + 9 + 13 +.....T_{n}\right] =435 \sqrt{3}$
$= \sqrt{3}\times\frac{n}{2}\left[2+\left(n-1\right)4\right] = 435\sqrt{3}$
$2n + 4n^{2} - 4n = 870
= 4n^{2} - 2n - 870 = 0$
$= 2n^{2} - n - 435 = 0$
$n = \frac{1\pm\sqrt{1+4\times2\times435}}{4}$
$= \frac{1\pm59}{4}$
$= \frac{1+59}{4} = 4; \frac{1-59}{4}$