Question

How many terms are there in the sequence $9,12,15,......66$?

• A. $18$
• B. $19$
• C. $20$
• D. $21$

Answer 1

Answer: (C)

$20$

Here first term $a=9$,common difference $d=12-9=3$ and $a_n=66$
Since, $a_n=a+(n-1)d$
$\Rightarrow 66=9+(n-1)3$
$\Rightarrow 57=(n-1)3$
$\Rightarrow 19=n-1$
$\therefore \Rightarrow n=20$
$\therefore$ Option C is correct.