# Ex.5.2 Q7 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

Find the \(31^\rm{st}\) term of an AP whose \(11^\rm{th}\) term is \(38\) and the \(16^\rm{th}\) term is \(73.\)

## Text Solution

**What is Known?**

\(11^\rm{th}\) and \(16^\rm{th}\) term of AP**. **

**What is Unknown?**

\(31^\rm{st}\) term of 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:**

\[\begin{array}{l}{a_n} = a + (n - 1)d\\{a_{11}} = 38\\a + (11 - 1)d = 38\\a + 10d = 38 \qquad \dots\left( 1 \right)\\{a_{16}} = 73\\a + 15d = 73\,\qquad \dots(2)\end{array}\]

By solving the two equations (1) & (2) for \(a,d\)

\[\begin{align}5d &= 35\\d &= 7\end{align}\]

Putting \(d\) in the (1) equation

\[\begin{align}a &= 38 - 70\\ &= - 32\end{align}\]

\(31^\rm{st}\) terms is,

\[\begin{align}{a_{31}} &= a + (31 - 1)d\\ &= - 32 + 30 \times 7\\ &= - 32 + 210\\& = 178\end{align}\]

The \(31^\rm{st}\) term of AP is \(178.\)