Question

Find the sum of all natural numbers lying between $100$ and $200$ which leave a remainder of $2$ when divided by $5$ in each case.

Answer 1

Hint: We will calculate the nearest value to $100$ which leaves remainder $2$ when divided by $5$ in each case and also the number nearest to $200$ which leaves remainder $2$ when divided by $5$ in each case. From both this values we can write a series of numbers which leaves remainder $2$ when divided by $5$ in each case by taking the lowest term as the first term of A.P and highest term as last term of A.P and common difference $5$. Now we have an A.P of numbers which leave a remainder of $2$ when divided by $5$ in each case. To find the sum of the numbers we will use the formula for sum of $n$ numbers of A.P i.e. ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.

Complete step-by-step answer:
The remainder when we divide $100$ with $5$ is zero. But we need remainder as $2$, so we will add $2$ to the $100$, then we will get $100+2=102$. So $102$ is the first number which leave a remainder of $2$ when divided by $5$ after the number $100$. So, we can take it as first term of A.P $\Rightarrow a=102$.

The remainder when we divide $200$ with $5$ is zero. But we need remainder as $2$, so we will subtract $5-2=3$ from $200$, then we will get $200-3=197$. So $197$ is the last number which leave a remainder of $2$ when divided by $5$ before the number $200$. So, we can take it as last term of A.P $\Rightarrow {{a}_{n}}=197$.
In A.P we have the formula for the ${{n}^{th}}$ term as ${{a}_{n}}=a+\left( n-1 \right)d$
We have ${{a}_{n}}=197$, $a=102$, $d=5$ , now the value of $n$ is given by
$\begin{align}
  & {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 197=102+\left( n-1 \right)5 \\
 & \Rightarrow 197-102=\left( n-1 \right)5 \\
 & \Rightarrow 95=\left( n-1 \right)5 \\
 & \Rightarrow n-1=19 \\
 & \Rightarrow n=20 \\
\end{align}$
Now the sum of the terms in A.P is given by
$\begin{align}
  & {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{20}{2}\left[ 2\times 102+\left( 20-1 \right)5 \right] \\
 & \Rightarrow {{S}_{n}}=10\left( 204+95 \right) \\
 & \Rightarrow {{S}_{n}}=2990 \\
\end{align}$
Hence the required sum of the numbers is $2990$.

Note: We can also find the number terms $100$ and $200$ which leave a remainder of $2$ when divided by $5$ in each case by subtracting the number of terms which leave a remainder of $2$ when divided by $5$ from $0$ to $100$ from the number of terms which leave a remainder of $2$ when divided by $5$ from $0$ to $200$. Mathematically,
$\begin{align}
  & n=\dfrac{200}{5}-\dfrac{100}{5} \\
 & \Rightarrow n=40-20 \\
 & \Rightarrow n=20 \\
\end{align}$
From both the methods we got the same result.