Question

Find the sum of all integers between 78 & 500 which is divisible by 7.

Answer 1

Hint: First of all we have to find the first & last term of the series. Then we will apply the formula of the terms that are in Arithmetic Progress.

Formula used:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\text{ or }{{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$
n number of terms l is the last term
a is first term
d is common difference (i.e. difference between any two consecutive terms i.e. ${{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}$
${{a}_{n}}=a+\left( n-1 \right)d$

Complete step-by-step answer:
We have to find numbers between 78 & 500 which are divisible by 7
By observing we can say 84 is the first term and last term is 497
84, 91, 98, ………. 497
$a=84\text{ }d=91-84=7\text{ }{{a}_{n}}=497$
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\
 &\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d \\
 &\Rightarrow 497=84+\left( n-1 \right)7 \\
 &\Rightarrow 497-84=\left( n-1 \right)7 \\
 &\Rightarrow 413=\left( n-1 \right)7 \\
\end{align}$
$\begin{align}
  & \dfrac{413}{7}=n-1 \\
 &\Rightarrow 59=n-1 \\
 &\Rightarrow 60=n \\
 &\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ a+l \right] \\
 &\therefore {{S}_{n}}=\dfrac{60}{2}\left[ 84+497 \right] \\
\end{align}$
$=30\left[ 581 \right]=17,430$

Additional information:
 We can also solve this question by using the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . Also the arithmetic progress is nothing but to add the terms which have a common difference i.e. the same number is added or subtracted in each successive term.

Note: By last term formula we find the value of $n$ then by substituting the values of first term, common difference and number of terms in the sum of term formula we get the desired result.