Question

Sum of 1 to n natural numbers is 36, then find the value of n .

Answer 1

It is given that,
a = 1
d = 1
S_n = 36

Now,
$$\begin{gathered} S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ \Rightarrow 36=\frac{n}{2}\left(2\left(1\right)+\left(n-1\right)\left(1\right)\right) \\ \Rightarrow 36\times 2=n\left(2+n-1\right) \\ \Rightarrow 72=n\left(n+1\right) \\ \Rightarrow n^2+n-72=0 \\ \Rightarrow n^2+9n-8n-72=0 \\ \Rightarrow n\left(n+9\right)-8\left(n+9\right)=0 \\ \Rightarrow \left(n+9\right)\left(n-8\right)=0 \\ \Rightarrow n=-9\mathrm{or}n=8 \\ \Rightarrow n=8\left(\because n\ne -9\right) \end{gathered}$$

Hence, the value of n is 8.