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

## Question

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

## 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 \({a_7} - {a_5} = 12\)

\[\begin{align} a+\left( 3-1 \right)d&=16 \\ a+2d&=16 \qquad \dots\text{Equation }\left( 1 \right) \\\end{align}\]

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 \(4, \,10,\, 16,\, 22,\,\dots.\)