The sums of n terms of three arithmetical progressions are $S_{1}, S_{2}$ and $S_{3}$ . The first term of each is unity and the common differences are $1, 2$ and $3$ respectively. Prove that $S_{1} + S_{3} = 2 S_{2}$ .

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Solution

Here $a = 1$ for all and $d = 1, 2, 3,$ respectively for $S_{1}, S_{2}, S_{3},$ and $n = n$ for all.
$S_{1} + S_{3} = (n / 2) [2.1 + (n - 1) .1] + (n / 2) [2.1 + (n - 1) .3]$
$= (n / 2) [n + 1] + (n / 2) [3 n - 1]$
$= (n / 2) 4 n = 2 n^{2}$
$2 S_{2} = 2. (n / 2) [2.1 + (n - 1) .2] = n . (2 n) = 2 n^{2}$
$∴ S_{1} + S_{3} = 2. S_{2}$ .

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The sums of n terms of three arithmetical progressions are S1,S2 and S3. The first term of each is unity and the common differences are 1,2 and 3 respectively. Prove that S1+S3=2S2.

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