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For the next two (02) items that follow:

Let

S_n

denote the sum of first n terms of an AP and

3S_n=S_{2n}

[NDA II 2014]

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Question : 76 of 79

Marks: +1, -0

\text{ What is } S_{3n}: S_{n} \text{ equal to? }

$4:1$
$6:1$
$8:1$
$10:1$

Solution:

3 \times \frac{n}{2}[2a+(n-1)d] = \frac{2n}{2}[2a+(2n-1)d]

\Rightarrow 6a+3nd-3d=4a+4nd-2d

\Rightarrow 2a=nd+d=d(n+1)

\frac{S_{3n}}{S_n} = \frac{ \frac{3n}{2}[2a+(3n-1)d] }{ \frac{n}{2}[2a+(n-1)d] } = \frac{ \frac{3n}{2}[d(n+1)+(3n-1)d] }{ \frac{n}{2}[d(n+1)+(n-1)d] }

= \frac{ \frac{3n}{2}[4nd] }{ \frac{n}{2}[2nd] } = \frac{6}{1}

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