Question

Find the sum of the first 15 multiples of 8.

Answer 1

Hint: First write the first few multiples of 8. We observe that they are in A.P. Find the first term and common difference of this A.P. Then find the sum of first 15 terms of this A.P. using the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.

Complete step-by-step answer:
In this question, we need to find the sum of the first 15 multiples of 8.
The first few multiples of 8 are 8, 16, 24, 32, …
We can observe that these multiples are in an A.P.
The first term of this A.P. is a = 8.
Let us find the common difference of this A.P.
The common difference, d = 16 – 8 = 24 – 16 = 8
Hence, for this A.P., a = 8 and d = 8.
We need to find the sum of the first fifteen terms of this A.P.
i.e. we need to find ${{S}_{15}}$, where ${{S}_{n}}$ denotes the sum of the first n terms of an A.P.
We know that the general formula to calculate the sum of the first n terms of an A.P. i.e. ${{S}_{n}}$is the product of half of the number of terms and the sum of twice of the first term to the product of the common difference and the number of terms minus one.
i.e. ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Substituting a = 8, d = 8, and n = 15 in the above equation, we will get the following:
${{S}_{15}}=\dfrac{15}{2}\left[ 2\times 8+\left( 15-1 \right)8 \right]$
${{S}_{15}}=\dfrac{15}{2}\left[ 16+14\times 8 \right]$
${{S}_{15}}=\dfrac{15\times 128}{2}=960$
So, the sum of the first 15 multiples of 8 is equal to 960.
This is the final answer.

Note: In this question, it is very important to observe and realize that the multiples of 8 form an A.P. Also, remember the formula to calculate the sum of the first n terms of an A.P. i.e. ${{S}_{n}}$is the product of half of the number of terms and the sum of twice of the first term to the product of the common difference and the number of terms minus one i.e. ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$