Question

For a sequence of natural numbers find $S_{50}-S_{40}$

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 $S_{50}$ with the first term is $a=1$, common difference is $d=2-1=1$ and $n=50$, therefore, the sum is:

$S_{50}=\frac{50}{2}\left[\right(2\times 1)+(50-1\left)\right(1\left)\right]=25(2+49)=25\times 51=1275$

Now, we find $S_{40}$ with the first term is $a=1$, common difference is $d=2-1=1$ and $n=40$, therefore, the sum is:

$S_{40}=\frac{40}{2}\left[\right(2\times 1)+(40-1\left)\right(1\left)\right]=20(2+39)=20\times 41=820$

Thus, $S_{50}-S_{40}=1275-820=455$

Hence, $S_{50}-S_{40}=455$.