Single Choice

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$$15$$
Correct Answer

Solution

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

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

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

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

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

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

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

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

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

$$=\displaystyle\frac{1+59}{4}=4; \frac{1-59}{4}$$

$$n=14.5,15$$

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