If the sum of the first n terms of the series | vedclass

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Question

$\sqrt 3 \left[ {1 + \sqrt {25}  + \sqrt {81}  + \sqrt {69}  + .....} \right] = 435\sqrt 3 $

$ \Rightarrow \sqrt 3  = \left[ {1

Vedclass · Accepted Answer

$15$

Vedclass · Answer

$\sqrt 3 \left[ {1 + \sqrt {25}  + \sqrt {81}  + \sqrt {69}  + .....} ight] = 435\sqrt 3 $

$ \Rightarrow \sqrt 3  = \left[ {1 + 5 + 9 + 13 + ..... + {T_n}} ight] = 435\sqrt 3 $

$ \Rightarrow \sqrt 3  	imes \frac{n}{2}\left[ {2 + \left( {n - 1} ight)4} ight] = 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 	imes 2 	imes 432}}{4} = \frac{{1 \pm 59}}{4}$

$	herefore \,\,\,\,n = \frac{{1 \pm 59}}{4} = 15$; or $ = \frac{{1 - 59}}{4} = 14.5$

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

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