Question

What is the sum of the first $100$ multiples of $12$ ?

Answer 1

Hint: The given problem requires us to find the sum of the first hundred multiples of twelve. Consecutive multiples of any number constitute an arithmetic progression since the difference between consecutive terms is constant. So, we will first write the initial terms in the series to find the common difference and the first term. For finding out the sum of an arithmetic progression, we need to know the first term, the common difference, and the number of terms in the arithmetic progression. We can find out the number of terms in an AP using the formula for general terms of an arithmetic progression.

Complete step-by-step solution:
So, we have to find the sum of the first $100$ multiples of $12$.
So, we write a few initial terms as: $12,24,36,48,...$
The difference between any two consecutive terms of the given series is constant. So, the given sequence is an arithmetic progression.
Now, we have to find the sum of this arithmetic progression.
Here, first term $ = a = 12$.
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common difference \[ = d = 24 - 12 = 12\]
So, $d = 12$ .
We also need to know the number of terms in the arithmetic progression in order to find the value of the sum of arithmetic progression.
Now, we know that we have to find the sum of the first hundred multiples of $12$. So, the number of terms in the AP is $12$.
Now, we can find the sum of the given arithmetic progression using the formula $S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Hence, the sum of AP $ = S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Substituting the values of a, d and n in the formula, we get,
$ \Rightarrow S = \dfrac{{100}}{2}\left[ {2\left( {12} \right) + \left( {100 - 1} \right)\left( {12} \right)} \right]$
Opening the brackets and cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow S = 50\left[ {24 + 99\left( {12} \right)} \right]$
Simplifying the expression, we get,
$ \Rightarrow S = 50\left[ {24 + 1188} \right]$
$ \Rightarrow S = 50 \times 1212$
Doing the calculations, we get,
$ \Rightarrow S = 60600$
So, the sum of the first $100$ multiples of $12$ is $60600$.

Note: Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. The sum of n terms of an arithmetic progression can be calculated if we know the first term, the number of terms and difference of the arithmetic series as: $S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.