Question
Find the sum of first n natural numbers.
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Answer
The first n natural numbers are 1, 2, 3, 4, 5, ..., n.
Here, a = 1 and d = (2 - 1) = 1
Sum of n terms of an AP is given by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2\times1+(\text{n}-1)\times1\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2+\text{n}-1\big]=\big(\frac{\text{n}}{2}\big)\times(\text{n}+1)=\frac{\text{n}(\text{n}+1)}{2}$
Here, a = 1 and d = (2 - 1) = 1
Sum of n terms of an AP is given by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2\times1+(\text{n}-1)\times1\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2+\text{n}-1\big]=\big(\frac{\text{n}}{2}\big)\times(\text{n}+1)=\frac{\text{n}(\text{n}+1)}{2}$
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