Question

For a sequence of natural numbers find $n$ if $S_n=55$

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 $S_n=55$ with the first term is $a=1$, common difference is $d=2-1=1$, therefore,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 55=\frac{n}{2}\left[\right(2\times 1)+(n-1\left)\right(1\left)\right]\Rightarrow 55=\frac{n}{2}(2+n-1)\Rightarrow 55=\frac{n(n+1)}{2}\Rightarrow 55\times 2=n^2+n\Rightarrow n^2+n=110\Rightarrow n^2+n-110=0\Rightarrow n^2+11n-10n-110=0\Rightarrow n(n+11)-10(n+11)=0\Rightarrow n+11=0,n-10=0\Rightarrow n=-11,n=10$

Thus, ignoring the negative value of $n$, we have $n=10$.

Hence, $n=10$.