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UPSC NDA II 2014 Mathematics Paper

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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 : 46 of 120

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

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

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

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

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

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