JEE Main On 8 April 2017 Question 8
Question: If the um of the first n terms of the series $ \sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+…….. $ is $ 435\sqrt{3}, $ then n equals. [JEE Online 08-04-2017]
Options:
A) 13
B) 15
C) 29
D) 18
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \sqrt{3}[1+\sqrt{25}+\sqrt{81}+\sqrt{69}+……]=435\sqrt{3} $
$ \sqrt{3}[1+5+9+13+…..T _{n}]=435\sqrt{3} $
$ =\sqrt{3}\times \frac{n}{2}[2+(n-1)4]=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\times 2\times 435}}{4} $
$ =\frac{1\pm 59}{4} $
$ =\frac{1+59}{4}=4;\frac{1-59}{4} $
$=-14.5$
n cannot be negative
so n=15
