Question

For what value of $n$, the $nth$ terms of the APs $63,65,67,.....$ and $3,10,17,......$ are equal ?

Answer 1

First A.P is

$63,65,67,...\dots$

first term of this A.P $a_1=63$

second term of this A.P $a_2=65$

common difference $d=a_2-a_1=65-63=2$

hence nth term of this A.P is given by

$a_n=a_1+(n-1)d$

$⟹a_n=63+(n-1)2$

$⟹a_n=63+2n-2$

$⟹a_n=61+2n...\dots .eq\left(1\right)$

second A.P is

$3,10,17,...\dots$

first term of this A.P $a_1^{\prime }=3$

second term of this A.P $a_2^{\prime }=10$

common difference $d^{\prime }=a_2^{\prime }-a_1^{\prime }=10-3=7$

hence nth term of this A.P is given by

$a_n^{\prime }=a_1^{\prime }+(n-1)d^{\prime }$

$⟹a_n^{\prime }=3+(n-1)7$

$⟹a_n^{\prime }=3+7n-7$

$⟹a_n^{\prime }=-4+7n...\dots .eq\left(2\right)$

we have to find that term $\left(n\right)$ for which the nth term of both A.P are equal

$a_n=a_n^{\prime }$

hence from eq(1) and eq(2)

$⟹61+2n=-4+7n$

$⟹7n-2n=61+4$

$5n=65$

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

$⟹n=13$

hence $13$th term of both the A.P's are equal.