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The sum of $n$ terms of three A.P’s are $s_{1}, s_{2}, s_{3}$ respectively. The first term of each is unity and common difference are 1, 2 and 3 respectively then $\frac{s_{1} + s_{3}}{s_{2}} =$

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answer is B.

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Detailed Solution

s_{1} = 1 + 2 + \dots, n terms = \frac{n (n + 1)}{2}

$s_{2} = 1 + 3 + 5 + \dots, n terms = n^{2}$

$s_{3} = 1 + 4 + 7 + \dots, n terms = \frac{n}{2} [2 + (n - 1) 3] = \frac{n}{2} [3 n - 1]$

$\frac{s_{1} + s_{3}}{s_{2}} = \frac{\frac{n^{2} + n}{2} + \frac{3 n^{2} - n}{2}}{n^{2}} = \frac{\frac{4 n^{2}}{2}}{n^{2}} = 2$

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