Question

How many terms are there in the sequence $3,6,9,12,....,111$?

Answer 1

Clearly the given sequence is an A.P. with first term $a=3$ and common difference $d=3$.

Let there be $n$ terms in the given sequence. Then,

$n^{th}$ term $=111$

$⟹a+(n-1)d=111$

$⟹3+(n-1)\times 3=111⟹n=37$

Thus, the given sequence contains $37$ terms.