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. $13$
• C. $29$
• D. $15$

Answer 1

Answer: (D)

$15$

$\Rightarrow$$\sqrt{3}[1+\sqrt{25}+\sqrt{81}+\sqrt{169}+......+T_n]=435\sqrt{3}$

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

$\Rightarrow Formula:S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$\Rightarrow$$\sqrt{3}\times \frac{n}{2}[2+(n-1\left)4\right]=435\sqrt{3}$

$\Rightarrow$$2n+4n^2-4n=870$

$\Rightarrow$$4n^2-2n-870=0$

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

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

$=\frac{1\pm 59}{4}$

$\Rightarrow \frac{1+59}{4}=15;\frac{1-59}{4}=-14.5$

$n$ cannot be negative

Hence, $n=15$