Sum of n terms of the series $$\sqrt 2 $$ $$ + $$ $$\sqrt 8 $$ $$ + $$ $$\sqrt {18} $$ $$ + $$ $$\sqrt {32} $$ $$ + $$ ....... is
A. $$\frac{{n\left( {n + 1} \right)}}{2}$$
B. $$2n\left( {n + 1} \right)$$
C. $$\frac{{n\left( {n + 1} \right)}}{{\sqrt 2 }}$$
D. $$1$$
Answer: Option C
Solution(By Examveda Team)
The series is given$$\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} $$ $$ + $$ ......
$$ \Rightarrow \sqrt 2 + 2\sqrt 2 + 3\sqrt 2 $$ $$ + $$ $$4\sqrt 2 $$ $$ + $$ ......
Here a = $$\sqrt 2 $$ and d = $$2\sqrt 2 $$ $$ - $$ $$\sqrt 2 $$ = $$\sqrt 2 $$
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2\sqrt 2 + \left( {n - 1} \right)\sqrt 2 } \right] \cr & = \frac{n}{2}\left[ {2\sqrt 2 + \sqrt 2 \,n - \sqrt 2 } \right] \cr & = \frac{n}{2}\left( {\sqrt 2 \,n + \sqrt 2 } \right) \cr & = \frac{{n\sqrt 2 }}{2}\left( {n + 1} \right) \cr & = \frac{{n\left( {n + 1} \right)}}{{\sqrt 2 }} \cr} $$
Related Questions on Progressions
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A. 5
B. 6
C. 4
D. 3
E. 7
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A. 600
B. 765
C. 640
D. 680
E. 690
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