# Ex.5.2 Q18 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

The sum of \(4^\rm{th}\) and \(8^\rm{th}\) terms of an A.P. is \(24\) and the sum of the \(6^\rm{th}\) and \(10^\rm{th}\) terms is \(44.\) Find the first three terms of the A.P.

## Text Solution

**What is Known:?**

Sum of \(4^\rm{th}\) and \(8^\rm{th}\) terms is \(24\) and sum of \(6^\rm{th}\) and \(10^\rm{th}\) terms is \(44.\)

**What is Unknown?**

First three terms of the AP.

**Reasoning:**

\({a_n} = a + \left( {n - 1} \right)d\) is the general term of AP. Where \({a_n}\) is the \(n\rm{th}\) term, \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

Let \(a\) be the first term and \(d\) the common difference.

Given,

\[\begin{align}{a_4} + {\rm{ }}{a_8}& = {\rm{ }}24\\\left( {a + 3d} \right) + \left( {a + 7d} \right) &= 24\\ \Rightarrow 2a + 10d &= 24\\ \Rightarrow a + 5d{\rm{ }} &= 12 \dots{\left( 1 \right)}\end{align}\]

Also,

\[\begin{align}{{a}_{6}}+{{a}_{10}}&=44 \\\left( a+5d \right)+\left( a+9d \right)&=44 \\\Rightarrow 2a+14d&=44 \\\Rightarrow a+7d&=22 \dots\left( 2 \right) \\\end{align}\]

On subtracting Equation (1) from (2), we obtain

\[\begin{align}(a + 7d)-\left( {a + 5d} \right) &= 22 - 12\\a + 7d - a - 5d &= 10\\2d &= 10\\d &= 5\end{align}\]

By Substituting the value of \(d = 5\) in Equation (1), we obtain

\[\begin{align}a + 5d &= 12\\a + 5 \times 5 &= 12\\a + 25 &= 12\\a &= - 13\end{align}\]

The first three terms are \(a\), \(\left( {a + d} \right)\) and \(\left( {a + 2d} \right)\)

Substituting the values of \(a\) and \(d\), we get \(–13, (–13+5)\) and \((–13+2×5)\)

i.e., \(–13, –8\) and \(–3\)

Therefore, the first three terms of this A.P. are \(−13, −8,\) and \(−3.\)