Question

Find the sum of all three-digit natural numbers, which are multiples of 7.

Answer 1

Hint: So here in this question First we will find the lowest and highest three-digit natural numbers which are multiples of 7. Then we will find the number of terms between these least and highest natural numbers which are multiples of 7 using the formula $l=a+\left( n-1 \right)d$. After that, we will find the sum of all such three-digit numbers which are multiples of 7 by using the formula $S=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.

Complete step-by-step solution:
We know that the least three-digit number which is a multiple of 7 is 105.
And the highest three-digit number which is a multiple of 7 is 994.
We know that that formula for $n^{th}$ term is given by $l=a+\left( n-1 \right)d …..... \left( i \right)$
where l is the nth term, a is the first term, n is the number of terms, and d is a common difference.
Here, a = 105, l = 994 and d = 7.
Putting these values in equation (i), we get
$\begin{align}
  & \text{ }994=105+\left( n-1 \right)7 \\
 & \Rightarrow 994-105=\left( n-1 \right)7 \\
 & \Rightarrow 889=\left( n-1 \right)7 \\
 & \Rightarrow n-1=127 \\
 & \Rightarrow n=128 \\
\end{align}$
We know that the sum of n terms in a sequence is given by
$S=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]................. \left( ii \right)$
Putting the values in equation (ii), we get
\[\begin{align}
  & S=\dfrac{128}{2}\left[ (2\times 105)+\left( 128-1 \right)7 \right] \\
 & =64\left[ 210+889 \right] \\
 & =64\times 1099 \\
 & =70,336
\end{align}\]
Hence, the sum of all three-digit natural numbers, which are multiples of 7 is 70336.

Note: Another formula for the sum of n terms is given by $S=\dfrac{n}{2}\left[ a+l \right]$ where l is the last term of the series. This formula can be easily used in questions like these where we know the last term of the sequence.