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

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## 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.

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 18

## 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 + 10{\rm{ }}d &= 24\\ \Rightarrow a + 5d{\rm{ }} &= 12\qquad{\rm{ }}\dots{\rm{Equation}\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 \qquad \dots\text{Equation}\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 $$−3.$$

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