Question

If the set of natural numbers is partitioned into subsets $S_1=\left\{1\right\},S_2=\left\{2,3\right\},S_3=\left\{4,5,6\right\}$ and so on.

Then the sum of the terms in $S_{50}$ is

• A. $62525$
• B. $25625$
• C. $62500$
• D. None of these

Answer 1

The correct option is A

$62525$

Explanation for the correct option

Step 1: Solve for the last term of set $S_{50}$

The given sets are $S_1=\left\{1\right\},S_2=\left\{2,3\right\},S_3=\left\{4,5,6\right\}$.

The last term of $S_1$ is $l_1=1$.

The last term of $S_2$ is $l_2=3=1+2$.

The last term of $S_3$ is $l_3=6=1+2+3$.

Thus, the last term of the set $S_n$ can be given by $l_n=\left(1+2+3+.....+n\right)$.

The sum of the first $n$ terms can be given by $\frac{n\left(n+1\right)}{2}$.

Thus, $l_n=\frac{n\left(n+1\right)}{2}$.

Therefore, the last term of the set $S_{50}$ can be given by $l_{50}=\frac{50\left(50+1\right)}{2}=1275$.

Step 2: Solve for the sum of terms in $S_{50}$

Therefore, the last term of the set $S_{49}$ can be given by $l_{49}=\frac{49\left(49+1\right)}{2}=1225$.

Thus, the first term of the set $S_{50}$ is $a_{50}=\left(l_{49}+1\right)=\left(1225+1\right)=1226$.

So, the set $S_{50}$ can be given by $S_{50}=\left\{1226,1227,......,1275\right\}$.

Thus, the number of terms, $n=50$.

It is known that, if the first term of an A.P. is $a$ and the last term of an A.P. is $l$, then the sum of $n$ terms can be given by $S_n=\frac{n\left(a+l\right)}{2}$.

So, the sum of the terms in $S_{50}$ can be given by $\frac{n}{2}\left(a_{50}+l_{50}\right)=\frac{50}{2}\left(1226+1275\right)=25\times 2501=62525$.

Therefore, the sum of the given series $S_{50}$ is $62525$.

Hence, option(A) i.e. $62525$ is the correct option.