Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively,
Question:

Let the sum of $n, 2 n, 3 n$ terms of an A.P. be $S_{1}, S_{2}$ and $S_{3}$, respectively, show that $S_{3}=3\left(S_{2}-S_{1}\right)$

Solution:

Let a and b be the first term and the common difference of the A.P. respectively.

Therefore,

$\mathrm{S}_{1}=\frac{n}{2}[2 a+(n-1) d]$ $\ldots(1)$

$\mathrm{S}_{2}=\frac{2 n}{2}[2 a+(2 n-1) d]=n[2 a+(2 n-1) d]$ $\ldots(2)$

$\mathrm{S}_{3}=\frac{3 n}{2}[2 a+(3 n-1) d]$ $\ldots(3)$

From (1) and (2), we obtain

$\mathrm{S}_{2}-\mathrm{S}_{1}=n[2 a+(2 n-1) d]-\frac{n}{2}[2 a+(n-1) d]$

$=n\left\{\frac{4 a+4 n d-2 d-2 a-n d+d}{2}\right\}$

$=n\left[\frac{2 a+3 n d-d}{2}\right]$

$=\frac{n}{2}[2 a+(3 n-1) d]$

$\therefore 3\left(\mathrm{~S}_{2}-\mathrm{S}_{1}\right)=\frac{3 n}{2}[2 a+(3 n-1) d]=\mathrm{S}_{3}$ [From (3)]

Hence, the given result is proved.

 

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