Question

Find $a_{30}$ given that the first few terms of an arithmetic sequence are given by $6,12,18,...$

Answer 1

Given series is $6,12,\mathrm{18...}$

In a arithmetic series $a_n=a+(n-1)d$, where $a$ is the first term of the sequence, $d$ is the common difference and $n$ is the number of the term to find.

Since the first term is $6$ and second term is $12$, therefore, the common difference is :

$d=12-6=6$ implies $d=6$

Now we find $a_{30}$ given that $a=6$, $d=6$ and $n=30$

$a_{30}=6+(30-1)\left(6\right)=6+174=180$