Test Index

UPSC NDA II 2014 Mathematics Paper

Show Para Hide Para

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]

© examsnet.com

Question : 47 of 120

Marks: +1, -0

What is

S_{3n}:S_{2n}

$2:1$
$3:1$
$4:1$
$5:1$

Solution:

\frac{S_{3n}}{S_{2n}} = \frac{6 n^{2} d}{3 n^{2} d}

[from Eqs. (ii) and (iii)]

= \frac{6}{3} = 2: 1

© examsnet.com

Go to Question:

Prev Question

Next Question