Question:
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)
Solution:
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.