If the sum of the first n terms of the series


\small \sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+......is 435\sqrt{3}

then n eqals :

  • Option 1)

    18

  • Option 2)

    15

  • Option 3)

    13

  • Option 4)

    29

 

Answers (2)

As we learnt in

Sum of n terms of an AP -

S_{n}= \frac{n}{2}\left [ 2a +\left ( n-1 \right )d\right ]

 

and

Sum of n terms of an AP

 

S_{n}= \frac{n}{2}\left [ a+l\right ]

- wherein

a\rightarrow first term

d\rightarrow common difference

n\rightarrow number of terms

 

 \sqrt{3} +\sqrt{75} +\sqrt{243} +\sqrt{507} .... =435\sqrt{3}

\Rightarrow\sqrt{3} +\sqrt{25\times3} +\sqrt{81\times3} +\sqrt{169\times3} +.....=435\sqrt{3}

\Rightarrow\sqrt{3} +5\sqrt{3} +9\sqrt{3} +13\sqrt{3}+ ....

\Rightarrow\sqrt{3} \left [ 1 + 5 + 9 + 13+ ...... \right ]

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

\Rightarrow \sqrt{3}\times{n}\left [ 2n - 1 \right ]=435\sqrt{3}

\Rightarrow \left ( 2n^{2}-n} \right ) = 435

\Rightarrow 2n^{2} - n - 435 = 0

\therefore\ \;n = \frac{1+\sqrt{3481}}{4} = \frac{1+59}{4} =\frac{60}{4}=15

 

 


Option 1)

18

Incorrect option    

Option 2)

15

Correct option

Option 3)

13

Incorrect option

Option 4)

29

Incorrect option

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