Question

Find the sum of the first 15 multiples of 8.

Answer 1

Hint: For solving this question we will simply write down the first 15 multiples of 8 and add then we will use the formula of the sum of $n$ terms which are in Arithmetic progression (A.P.) directly to get the correct answer.

Complete step by step solution:
Given:
We have to find the sum of the first 15 multiples of 8.
Now, the first 15 multiples of 8 are given below:
$\begin{align}
  & 8\times 1=8 \\
 & 8\times 2=16 \\
 & 8\times 3=24 \\
 & 8\times 4=32 \\
 & 8\times 5=40 \\
 & 8\times 6=48 \\
 & 8\times 7=56 \\
 & 8\times 8=64 \\
 & 8\times 9=72 \\
 & 8\times 10=80 \\
 & 8\times 11=88 \\
 & 8\times 12=96 \\
 & 8\times 13=104 \\
 & 8\times 14=112 \\
 & 8\times 15=120 \\
\end{align}$

Now, the first 15 multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120. As the difference between any two-consecutive term in the sequence 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120 is 8. So, the sequence 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120 will be an Arithmetic progression.
Now, we will use the following formulae directly to solve this question.
If, $a$ is the first term and $l$ is the ${{n}^{th}}$ term of an A.P. of $n$ terms. Then,
Sum of first $n$ terms of A.P. $=\dfrac{n}{2}\left( a+l \right)$
Now, for the sum of the first 15 multiples of 8. We can apply the above formula directly with the following value:
$\begin{align}
  & n=15 \\
 & a=8 \\
 & l=120 \\
\end{align}$
Then, the sum of the first 15 multiples of 8 $=\dfrac{15}{2}\times \left( 8+120 \right)=\dfrac{15}{2}\times 128=960$.

Thus, the sum of the first 15 multiples of 8 is 960.

Note: The problem is very easy to solve but the student should be careful while putting the values of different variables in the formulae to get the correct answer without any mistake. If one doesn’t know the formula of summation of $n$ terms of A.P. then we can simply add the first 15 multiples of 8 directly to get the correct answer.