# Ex.5.2 Q15 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

For what value of n, are the nth terms of two APs \(63,\, 65,\, 67,\,\)and \(3,\, 10,\, 17,\,\dots\) equal?

## Text Solution

**What is Known:?**

Two different APs

**What is Unknown?**

Value of \(n\) so that two APs have equal \(n^\rm{th}\) 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:**

If \(n\rm{th}\) terms of the two APs \(63, \,65,\, 67,\dots\) and \(3,\, 10,\, 17,\, \dots\) are equal.

Then, \(63 + \left( {n--1} \right)2 = 3 + \left( {n--1} \right)7 \qquad \text{Equation (1)}\)

[Since In \(1\rm{st}\) AP, \(a = 63\), \(d = 65--3 = 2\) and in \(2\rm{nd}\) AP , \(a = 3\),\(d = 10--3 = 7\)]

By Simplifying Equation (1)

\[\begin{align}7\left( {n-1} \right)-2\left( {n-1} \right)&= 63-3\\7n - 7 - 2n + 2 &= 60\\5n - 5 &= 60\\n &= \frac{{65}}{5}\\n &= 13\end{align}\]

The \(13^\rm{th}\) terms of the two given APs are equal.