Question
Find the sum of first n natural numbers.

Answer

A.P. formed is 1, 2, 3, 4, ..., n. Here, $\text{a}=1$ $\text{d}=1$ $\text{l}=\text{n}$ So sum of n terms $=\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $=\frac{\text{n}}{2}[2+(\text{n}-1)1]$ $=\frac{\text{n}(\text{n}+1)}{2}$ is the sum of first n natural numbers.

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