Answer
Explicit Formula: $a_{n}=4$$(0.1)^{n-1}$
First five terms: $4,0.4,0.04,0.004,0.0004$
Work Step by Step
Recall the explicit formula for a geometric sequence:
$a_{n}=$ $a_{1}$ $\times$ $r^{n-1}$
where
$a_n$ = $n^{\text{th}}$ term;
$a_1$ = first term;
$n$ = term number
$r$ = common ratio
The given quadratic sequence has:
$a_{1}=4$ and $r=0.1$
Thus, substituting these values into the formula abov gives:
$a_{n}=4$$(0.1)^{n-1}$
The first five terms are:
$a_{1}=4$
$a_{2}=4$$(0.1)^{2-1}$$=0.4$
$a_{3}=4$$(0.1)^{3-1}$$=0.04$
$a_{4}=4$$(0.1)^{4-1}$$=0.004$
$a_{5}=4$$(0.1)^{5-1}$$=0.0004$
