Question

For a sequence of natural numbers find $n$ if $S_n=15$

Answer 1

We know the sum of $n$ terms of an A.P with first term $a$ and the common difference $d$ is:

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

Since the sequence is of natural numbers.

We are given $S_n=15$ with the first term is $a=1$, common difference is $d=2-1=1$, therefore,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 15=\frac{n}{2}\left[\right(2\times 1)+(n-1\left)\right(1\left)\right]\Rightarrow 15=\frac{n}{2}(2+n-1)\Rightarrow 15=\frac{n(n+1)}{2}\Rightarrow 15\times 2=n^2+n\Rightarrow n^2+n=30\Rightarrow n^2+n-30=0\Rightarrow n^2+6n-5n-30=0\Rightarrow n(n+6)-5(n+6)=0\Rightarrow n+6=0,n-5=0\Rightarrow n=-6,n=5$

Thus, ignoring the negative value of $n$, we have $n=5$.

Hence, $n=5$.