# Ex.5.3 Q14 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

Find the sum of the odd numbers between \(0\) and \(50.\)

## Text Solution

**What is Known?**

Odd numbers between \(0\) and \(50\)

**What is Unknown?**

Sum of the odd numbers between \(0\) and \(50\)

**Reasoning:**

Sum of the first

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

**Steps:**

The odd numbers between \(0\) and \(50\) are \(1, \,3,\, 5,\, 7,\, 9 \,... \,49\)

Therefore, it can be observed that these odd numbers are in an A.P.

Hence,

- First term, \(a = 1\)
- Common difference, \(d = 2\)
- Last term, \(l = 1\)

We know that nth term of AP, \(\,{a_n} = l = a + \left( {n - 1} \right)d\)

\[\begin{align}49 &= 1 + \left( {n - 1} \right)2\\48& = 2\left( {n - 1} \right)\\n - 1 &= 24\\n &= 25

\end{align}\]

We know that sum of \(n\) terms of AP,

\[\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\{S_{25}} &= \frac{{25}}{2}\left( {1 + 49} \right)\\ &= \frac{{25}}{2} \times 50\\& = 25 \times 25\\ &= 625\end{align}\]