Question

The sum of first five multiples of 3 is
A. 45
B. 55
C. 65
D. 75

Answer 1

Hint: To solve this question, we need to have the knowledge of arithmetic progression. For example, let us consider an arithmetic progression a, a+d, a+2d, …….., a+(n-1)d having n terms in AP, then the sum of these n terms can be represented as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n- \right)d \right]$.

Complete step-by-step answer:
In this question, we are asked to find out the sum of the first five multiples of 3. We know that the first five multiples of 3 are 3, 6, 9, 12 and 15, which forms an arithmetic progression with 3 as the first term of the AP and (6 – 3) = 3 as the common difference of the AP. So, to find the first five terms of an AP, we need to know that, sum of n terms of an AP can be written as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n- \right)d \right]$. Where $n$ represents the number of terms in the AP, $a$ represents the first term of the AP and $d$ represents the common difference of the AP. So, from the question, we can write that, $n=5,a=3$ and $d=3$.
Therefore, the sum of the first five multiples of 3 can be written as:
 $\begin{align}
  & {{S}_{5}}=\dfrac{5}{2}\left[ 2\left( 3 \right)+\left( 5-1 \right)\left( 3 \right) \right] \\
 & \Rightarrow \dfrac{5}{2}\left[ 6+4\left( 3 \right) \right] \\
 & \Rightarrow \dfrac{5}{2}\left[ 6+12 \right] \\
 & \Rightarrow \dfrac{5}{2}\left[ 18 \right] \\
 & \Rightarrow 5\times 9 \\
 & \Rightarrow 45 \\
\end{align}$
Therefore, we can write that the sum of the first five multiples of 3 is 45.
Hence, option (A) is the correct option.

Note: In this question, we can also find the sum of the first five multiples of 3 by simply adding them. But if we get a larger number like 12, then the solution will get lengthier and it will be very time consuming. So, it is better to apply the formula of finding the sum of $n$ terms of AP, that is, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n- \right)d \right]$.