# Ex.5.3 Q13 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

Find the sum of first \(15\) multiples of \(8.\)

## Text Solution

**What is Known?**

Multiples of \(8\)

**What is Unknown?**

Sum of first \(15\) multiples of \(8,\) \({S_{15}}\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

The multiples of \(8\) are \(8, 16, 24, 32, \dots\)

These are in an A.P.,

Hence,

- First term, \(a = 8\)
- Common difference, \(d = 8\)
- Number of terms, \(n = 15\)

As we know that Sum of \(n\) terms,

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{15}} &= \frac{{15}}{2}\left[ {2 \times 8 + \left( {15 - 1} \right)8} \right]\\ &= \frac{{15}}{2}\left[ {16 + 14 \times 8} \right]\\

&= \frac{{15}}{2}\left[ {16 + 112} \right]\\& = \frac{{15}}{2} \times 128\\ &= 15 \times 64\\& = 960\end{align}\]