Question

For a sequence of natural numbers find $S_{30}+S_{15}$

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 find $S_{30}$ with the first term is $a=1$, common difference is $d=2-1=1$ and $n=30$, therefore, the sum is:

$S_{30}=\frac{30}{2}\left[\right(2\times 1)+(30-1\left)\right(1\left)\right]=15(2+29)=15\times 31=465$

Now, we find $S_{15}$ with the first term is $a=1$, common difference is $d=2-1=1$ and $n=15$, therefore, the sum is:

$S_n=\frac{15}{2}\left[\right(2\times 1)+(15-1\left)\right(1\left)\right]=\frac{15}{2}(2+14)=\frac{15\times 16}{2}=15\times 8=120$

Thus, $S_{30}+S_{15}=465+120=585$

Hence, $S_{30}+S_{15}=585$.