Question

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

Answer 1

The first number that is multiple of 3 is 3. So, the first term of the progression is 3 and the total number of multiples of 3 is 5. So the number of terms of the progression is 5. The common difference is 3. Now, apply the formula ${a_n} = {a_1} + \left( {n - 1} \right)d$ to get the last term of the progression. Then find the sum of the progression by using the formula ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$.

Formula used:
The general term of the arithmetic progression is given by,
${a_n} = a + \left( {n - 1} \right)d$
The sum of the progression is given by,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
where, ${a_n}$ is the last term.
${a_1}$ is the first term.
n is the number of terms.
d is a common difference.
${S_n}$ is the sum of the series.

Complete step-by-step answer:
Given:- First term, a= 3
Number of terms, n= 5
Common difference, d= 5
Now, find the last term of the series,
${a_5} = 3 + \left( {5 - 1} \right) \times 3$
Subtract 1 from 5 and multiply the result by 3,
${a_5} = 3 + 12$
Now add the terms of the right side,
${a_5} = 15$
Now use the summation formula to get the sum,
${S_5} = \dfrac{5}{2}\left( {3 + 15} \right)$
Add the terms in the brackets,
${S_5} = \dfrac{5}{2} \times 18$
Cancel out the common factors from both numerator and denominator and multiply the terms to get them,
${S_5} = 45$

Hence, the sum of the first five multiples of 3 is 45.

Note: This question can be done in another way also.
Let the sum of the first five multiples be S.
Then, the sum will be
$S = 3 + 6 + 9 + 12 + 15$
Take 3 common from the right side of the equation,
$S = 3\left( {1 + 2 + 3 + 4 + 5} \right)$
Now, apply the formula for the sum of n natural numbers ${S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}$. Put $n = 5$,
$S = 3 \times \dfrac{{5\left( {5 + 1} \right)}}{3}$
Add the terms in the bracket,
$S = 3 \times \dfrac{{5 \times 6}}{2}$
Cancel out the common terms and multiply,
$S = 45$
Hence, the sum of the first five multiples of 3 is 45.