Question

Find value of $n$ whose $n^{th}$ terms are equal of the AP $3,11,19$ and $7,11,15$

Answer 1

First sequence$:3,11,19,...$

$a=3,d=11-3=8$

$t_n=3+(n-1)8=3+8n-8=8n-5$

Second sequence$:7,11,15,...$

$a=7,d=11-7=4$

$t_n=7+(n-1)4=7+4n-4=4n+3$

Given that $n^{th}$ term of both the sequences are equal.

$\Rightarrow 8n-5=4n+3$

$\Rightarrow 8n-4n=5+3$

$\Rightarrow 4n=8$

$\therefore n=\frac{8}{4}=2$