Question

The $n^{th}$terms of the two A.P.'s $63,65,67,......$ and $3,10,17,.......$ are equal. Find $n.$

Answer 1

Step 1: Finding the common difference of the two A.P.

Given two A.P.'s are $63,65,67,......$ and $3,10,17,.......$

The common difference of A.P. $={(n+1)}^{th}term-n^{th}term$

The common difference for the A.P. $63,65,67,......$ is $65-63$

$=2$

The common difference for the A.P. $3,10,17,.......$ is $10-3$

$=7$

Step 2: Finding the $n^{th}$ term of the A.P.

The $n^{th}$ term of A.P. is $a+(n-1)d$.

where $a$ is the first term of A.P. and $d$ is the common difference pf A.P.

The $n^{th}$ term of the A.P. $63,65,67,......$ is $63+(n-1)\times 2$

$$\begin{gathered} =63+2n-2 \\ =61+2n \end{gathered}$$

The $n^{th}$ term of the A.P. $3,10,17,.......$ is $3+(n-1)\times 7$

$$\begin{gathered} =3+7n-7 \\ =7n-4 \end{gathered}$$

Step 3: Equating the $n^{th}$ terms of both the A.P.

The $n^{th}$ term of the A.P. $63,65,67,......=$The $n^{th}$ term of the A.P. $3,10,17,.......$

$\Rightarrow 61+2n=7n-4$

$\Rightarrow 7n-2n=61+4$

$\Rightarrow 5n=65$

$\Rightarrow n=\frac{65}{5}$

$\Rightarrow n=13$

Hence, the value of $n$ is $13$.