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8. Sequences and Series
hard
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
$29$
(JEE MAIN-2017)
Solution
$\sqrt 3 \left[ {1 + \sqrt {25} + \sqrt {81} + \sqrt {69} + …..} \right] = 435\sqrt 3 $
$ \Rightarrow \sqrt 3 = \left[ {1 + 5 + 9 + 13 + ….. + {T_n}} \right] = 435\sqrt 3 $
$ \Rightarrow \sqrt 3 \times \frac{n}{2}\left[ {2 + \left( {n – 1} \right)4} \right] = 435\sqrt 3 $
$ \Rightarrow 2n + 4{n^2} – 4n = 870$
$ \Rightarrow 4{n^2} – 2n – 870 = 0$
$ \Rightarrow 2{n^2} – n – 435 = 0$
$ \Rightarrow n = \frac{{1 \pm \sqrt 1 + 4 \times 2 \times 432}}{4} = \frac{{1 \pm 59}}{4}$
$\therefore \,\,\,\,n = \frac{{1 \pm 59}}{4} = 15$; or $ = \frac{{1 – 59}}{4} = 14.5$
Standard 11
Mathematics
