Question

For what value of $n$, are the $n^{th}$ terms of two APs $63,65,67$, and $3,10,17,\dots$ equal?

Answer 1

Step 1: Equation for first AP

Considering the first AP,

$63,65,67,\dots$

First term, $a=63$

Common difference,

$d=a_2-a_1$

$=65-63$

$=2$

We know, $n^{th}$ term of this A.P.

$a_n=a+(n-1)d$

$=63+(n-1)2$

$=63+2n-2$

$a_n=61+2n\dots \dots \dots \dots \dots \dots .\left(i\right)$

Step 2: Equation for second AP

Considering the second AP,

$3,10,17,\dots$

First term, $a=3$

Common difference,

$d=a_2-a_1$

$=10-3$

$=7$

We know, $n^{th}$ term of this A.P.

$a_n=a+(n-1)d$

$=3+(n-1)7$

$=3+7n-7$

$a_n=-4+7n\dots \dots \dots \dots \dots \dots .\left(i\right)$

Step 3: Calculate $n$

Given, $n^{th}$ term of these A.P.s are equal to each other.

Equating both these equations, we get,

$61+2n=7n-4$

$\Rightarrow$$61+4=5n$

$\Rightarrow$ $5n=65$

$\Rightarrow$ $n=13$

Hence, for $n=13$, the $n^{th}$ terms of the two APs are equal.