Question

Find the sum of the first 20 multiples of 13.
A. 2,720
B. 2,730
C. 2,740
D. 2,750

Answer 1

Hint: In this problem, we need to find the type of series in which the multiples of the 13 are present. Now, use the formula for the sum of \[n\] terms of A.P. to find the first sum of the first 20 multiples of 13.

Complete step-by-step answer:
The multiples of the 13 are shown below.
\[13,26,39,52, \ldots \ldots\]
It can be observed that the multiples of the 13 are in an arithmetic series whose first term is 13 and common difference is also 13.
Now, the formula for the sum of \[n\] term in arithmetic progression is shown below.
\[{S_n} = \dfrac{n}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right)\]
Here, \[a\] is the first term and \[d\] is a common difference.
Substitute 20 for\[n\], 13 for \[a\] and 13 for \[d\] in equation (1) to obtain the sum of the first 20 multiples of 13.
\[
  \,\,\,\,\,\,{S_{20}} = \dfrac{{20}}{2}\left\{ {2\left( {13} \right) + \left( {20 - 1} \right)\left( {13} \right)} \right\} \\
\Rightarrow {S_{20}} = 10\left\{ {26 + \left( {19} \right)\left( {13} \right)} \right\} \\ \Rightarrow {S_{20}} = 10\left\{ {26 + 247} \right\} \\
 \Rightarrow {S_{20}} = 10\left\{ {273} \right\} \\
 \Rightarrow {S_{20}} = 2730 \\
\]
Thus, the sum of the first 20 multiples of 13 is 2730; hence option (B) is the correct answer.

Note: In arithmetic progression the difference of the two successive numbers are the same or a constant value. In other words in an arithmetic sequence, the numbers differ from each other by common difference.