If S_n denotes the sum of n terms of an arith | vedclass

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(C) The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}\{2a + (n - 1)d\}$.We need to find the value of $(S_{2n} - S_n)$.$

Vedclass · Accepted Answer

$\frac{1}{3}S_{3n}$

Vedclass · Answer

(C) The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}\{2a + (n - 1)d\}$.We need to find the value of $(S_{2n} - S_n)$.$S_{2n} - S_n = \frac{2n}{2}\{2a + (2n - 1)d\} - \frac{n}{2}\{2a + (n - 1)d\}$$= n\{2a + 2nd - d\} - \frac{n}{2}\{2a + nd - d\}$$= \frac{n}{2}\{4a + 4nd - 2d - 2a - nd + d\}$$= \frac{n}{2}\{2a + 3nd - d\} = \frac{n}{2}\{2a + (3n - 1)d\}$Now,consider $S_{3n} = \frac{3n}{2}\{2a + (3n - 1)d\}$.Therefore,$\frac{1}{3}S_{3n} = \frac{1}{3} \cdot \frac{3n}{2}\{2a + (3n - 1)d\} = \frac{n}{2}\{2a + (3n - 1)d\}$.Thus,$(S_{2n} - S_n) = \frac{1}{3}S_{3n}$.

If $S_n$ denotes the sum of $n$ terms of an arithmetic progression,then the value of $(S_{2n} - S_n)$ is equal to

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