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# Ex.5.2 Q16 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

Determine the A.P. whose third term is $$16$$ and the $$7^\rm{th}$$ term exceeds the $$5^\rm{th}$$ term by $$12.$$

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 16

## Text Solution

What is Known:?

Third term and relation between $$5^\rm{th}$$ term and $$7^\rm{th}$$ term.

What is Unknown?

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.

Hence from given, $${a_3} = 16$$ and $${a_7} - {a_5} = 12$$

\begin{align} a+\left( 3-1 \right)d&=16 \\ a+2d&=16 \quad \dots\left( 1 \right) \\\end{align}

Using $${a_7} - {a_5} = 12$$

\begin{align}\left[ {a + \left( {7 - 1} \right)d} \right] - \left[ {a + \left( {5 - 1} \right)d} \right] &= 12\\ [a + 6d] - [a + 4d] &= 12\\2d &= 12\\d &= 6\end{align}

By Substituting this in Equation (1), we obtain

\begin{align}a + 2 \times 6 &= 16\\a + 12 &= 16\\a &= 4\end{align}

Therefore, A.P. will be $$4+6,\,\,4+2\times 6,\,\,4+3\times 6\,\dots$$

Hence the series will be $$4, \,10,\, 16,\, 22,\,\dots.$$

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