Question

The sum of the integers from 1 to 100 which are divisible by 3 and 5 is:
1. 2317
2. 2632
3. 315
4. 2489

Answer 1

Hint: Here we will write the series of the integers that are divisible by 3, divisible by 5 and divisible by 15. This series will form an arithmetic progression. We will use the formula for the ${{n}^{th}}$ term of an A.P and the formula for the sum of terms of an A.P to find the sum of the three A.P. Finally we will use these sums to find the sum of the integers from 1 to 100 that are divisible by either 3 or 5.

Complete step-by-step solution:
According to the question it is asked to us to find the sum of the integers from 1 to 100 which are divisible by 3 and 5. For finding this we will use the following formulas. The ${{n}^{th}}$ term of an A.P is given by the formula ${{a}_{n}}=a+\left( n-1 \right)d$, where $a$ is the first term of the A.P. and $d$ is the common difference. The sum of $n$ terms of an A.P is given by the formula ${{S}_{n}}=\left( \dfrac{n}{2} \right)\left( a+l \right)$, where $a$ is the first term of the A.P. and $l$ is the last term of the A.P.
We know that the LCM of 3 and 5 is 15. The numbers between 1 and 100 that are divisible by both 3 and 5 are 15, 30, 45, ……, 90.
So, here we can write as,
$\begin{align}
  & a=15 \\
 & {{a}_{n}}=90 \\
 & d=15 \\
 & \Rightarrow a+\left( n-1 \right)d=90 \\
 & \Rightarrow 15+\left( n-1 \right)15=90 \\
 & \Rightarrow n=\dfrac{90}{15}=6 \\
\end{align}$
Now we will find the sum. So,
 $\begin{align}
  & {{S}_{n}}=\left( \dfrac{n}{2} \right)\left( a+l \right) \\
 & \Rightarrow {{S}_{n}}=\left( \dfrac{6}{2} \right)\left( 15+90 \right) \\
 & \Rightarrow {{S}_{n}}=3\times \left( 105 \right) \\
 & \Rightarrow {{S}_{n}}=315 \\
\end{align}$
Hence the correct answer is option 3.

Note: In this sort of questions, we must be careful while choosing the first and last terms of the arithmetic progression. If the in-between terms are used in the question, then the first and last terms should not be included in the analysis.