Question

Show that the sequence $9,12,15,18,...$ is an A.P. Find its ${16}^{th}$ term and the general term.

Answer 1

We have,
$(12-9)=(15-12)=(18-15)=3$

Therefore, the given sequence is an A.P. with common difference $3$.

$a=$First term $=9$

$\therefore {16}^{th}$ term $=a_{16}=a+(16-1)d=a+15d$ $[\because a_n=a+(n-1\left)d\right]$
$⟹a_{16}=9+15\times 3=54$
$\because$ General term $=n^{th}$ term $=a+(n-1)d$
$\therefore a_n=9+(n-1)\times 3=3n+6$