Question

Find the sum of first $10$ multiples of $6$.

Answer 1

We have to find the sum of the first ${\text{n}}$ multiple of the number by using the arithmetic progression.
Arithmetic progressions is a sequence where each new term after the first is obtained by adding a constant ${\text{d}}$, called the common difference, to the preceding term. If the first term of the sequence is ${\text{a}}$ then the arithmetic progression is ${\text{a,a + d,a + 2d,a + 3d,}}....\;$ where the ${{\text{n}}^{{\text{th}}}}$ term is ${\text{a + }}\left( {{\text{n - 1}}} \right){\text{d}}$ .

Formula used: The sum of the terms of an arithmetic progression gives an arithmetic series. If we know the first value ${\text{a}}$ and the value of the last term ${\text{l}}$ instead of the common difference ${\text{d}}$ then, we can write the sum as
${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}\left( {{\text{a + l}}} \right)$ .

Complete step-by-step answer:
The first ten multiples of six are $6,12,18,24,30,36,42,48,54,60$ .
Clearly, this is an arithmetic series. Because the difference between the terms is a constant value $6$and we also know that the first term, ${\text{a}} = 6$ and the last term, ${\text{l}} = 60$ .
From the given, we know that, ${\text{n = 10}}$.
Now, we can use the formula for the sum of an arithmetic progression.
If the last term is given in the arithmetic series, then the sum of the first n terms be
$\Rightarrow {{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}\left( {{\text{a + l}}} \right)$ .
Now, for the sum of the first $10$ terms be
$\Rightarrow {{\text{S}}_{10}}{\text{ = }}\dfrac{{10}}{2}\left( {{\text{6 + 60}}} \right)$
By adding and dividing the above terms, we get
$\Rightarrow {{\text{S}}_{10}}{\text{ = 5}}\left( {66} \right)$
$\Rightarrow {{\text{S}}_{{\text{10}}}}{\text{ = 33}}0$

Therefore, the sum of the first $10$ multiples of $6$ is $330$.

Note: There is an alternative method for the given question, by using the following formula.
The sum of the terms of an arithmetic progression gives an arithmetic series. If the starting value is $a$ and the common difference is $d$ then the sum of the first ${\text{n}}$ terms is
${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a + }}\left( {{\text{n - 1}}} \right){\text{d}}} \right]$ .
From the given arithmetic series, we get ${\text{a = 6,d = 6 and n = 10}}$. Then substitute in the above formula we get,
$\Rightarrow {{\text{S}}_{10}}{\text{ = }}\dfrac{{{\text{10}}}}{{\text{2}}}\left[ {{\text{2}}\left( 6 \right){\text{ + }}\left( {{\text{10 - 1}}} \right)6} \right]$
Simplifying the terms we get,
$\Rightarrow {{\text{S}}_{10}}{\text{ = 5}}\left[ {{\text{12 + }}\left( 9 \right)6} \right]$
Multiplying the terms we get,
$\Rightarrow {{\text{S}}_{10}}{\text{ = 5}}\left[ {{\text{12 + 54}}} \right]$
Adding the terms,
$\Rightarrow {{\text{S}}_{10}}{\text{ = 5}}\left[ {66} \right]$
Hence we get,
$\Rightarrow {S_{10}} = 330$.
Hence, the sum of the first $10$ multiples of $6$ is $330$.