Question

Find the sum of the first 15 multiples of 8.

Answer 1

Hint: A series is said to be in an arithmetic sequence in which the difference between each consecutive term of the sequence is constant. In general, the AP series is represented by the formula\[{t_n} = a + \left( {n - 1} \right)d\] where $n$ the number of terms.
The behaviour of the series depends on the common difference, if it is positive then, the sequence is increasing towards the infinity, and if the difference is negative, then the series is decreasing to the negative infinity.
Here, in the question we need to determine the sum of the first 15 multiples of 8 which will be a series in an AP with the first term 8 and the common difference of 8 as well.


Complete step by step solution:
First 15 multiples of 8 follow the arithmetic progression series as: $8,16,24,32,.......$
Here, the first term is 8 i.e., $a = 8$ and the common difference is also 8 i.e., $d = 8$
The sum of the AP series with n terms is given as: $S = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
Here, the total number of terms in the series is 15 i.e., $n = 15$
Substitute a=8, d=8 and n=15 in the formula $S = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$ to determine the sum as:
$
  S = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] \\
   = \dfrac{{15}}{2}\left[ {2(8) + (15 - 1)(8)} \right] \\
   = \dfrac{{15}}{2}\left[ {16 + 14(8)} \right] \\
   = \dfrac{{15}}{2}\left[ {16 + 112} \right] \\
   = \dfrac{{15}}{2} \times 128 \\
   = 15 \times 64 \\
   = 960 \\
 $

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


Note: While solving the question, we must be aware of the fact that the series is given in which form i.e., in AP or in GP. This can be done by checking its common difference or common ratio.