Question

Find the number of terms in the sequence$:4,12,20,...108$

• A. $12$
• B. $19$
• C. $13$
• D. $14$

Answer 1

The correct option is D $14$

Given sequence: $4,12,20,...108$

Here first term $a=4$,

common difference $d=12-4=8$

and last term $a_n=108$

Since, $a_n=a+(n-1)d$
$\Rightarrow 108=4+(n-1)8$
$\Rightarrow 104=8n-8$
$\Rightarrow 112=8n$
$\Rightarrow n=14$

So, the number of terms in the given sequence is $14$