# Ex.5.2 Q8 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

An AP consists of $$50$$ terms of which $$3^\rm{rd}$$ term is $$12$$ and the last term is $$106.$$ Find the $$29^\rm{th}$$ term.

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 8

## Text Solution

What is Known:?

Number of terms in the AP.  $$3^\rm{rd}$$ term is $$12$$ and last term is $$106$$

What is Unknown?

$$29^\rm{th}$$ term.

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:

${a_n} = a + \left( {n - 1} \right)d$

Third term of AP is $$a + 2d$$

$a + 2d = 12 \qquad \dots(1)$

Last term $$= 106$$

$$50\rm{th}$$ term $$=106$$

\begin{align}a + \left( {50 - 1} \right)d &= 106\\a + 49d &= 106 \qquad \dots\left( 2 \right)\end{align}

Solving (1) & (2) for the values of $$a$$ and $$d$$.

\begin{align}47d &= 94\\d &= 2\end{align}

Putting $$d = 2$$ in equation (1)

\begin{align}a + 2 \times 2 &= 12\\a + 4 &= 12\\a &= 12 - 4\\a &= 8\end{align}

$$29^\rm{th}$$ term of AP is

\begin{align}{a_{29}} &= a + (29 - 1)d\\{a_{29}} &= 8 + \left( {28} \right)2\\{a_{29}} &= 8 + 56\\{a_{29}}& = 64\end{align}

$$29^\rm{th}$$ term of AP is $$64.$$

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