Solve the following


If S1 be the sum of (2n + 1) terms of an A.P. and S2 be the sum of its odd terms, the prove that:

S1 : S2 = (2n + 1) : (n + 1)


Let the A.P. be $a, a+\mathrm{d}, a+2 d \ldots$

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

$\Rightarrow S_{1}=\frac{2 n+1}{2}[2 a+(2 n) d]$

$\Rightarrow S_{1}=(2 n+1)(a+n d) \quad \ldots(i)$

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

$\Rightarrow S_{2}=\frac{n+1}{2}[2 a+2 n d]$

$\Rightarrow S_{2}=(n+1)[a+n d]$    ....(ii)

From (i) and (ii), we get:

$\frac{S_{1}}{S_{2}}=\frac{2 n+1}{n+1}$

Hence, proved.

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