These are the first six terms of a sequence with a = 2:

Mathematics

1 answer:

inn [45]3 years ago

7 0

Answer:

$a_{n}$ = 9 $a_{n-1}$

Step-by-step explanation:

Note there is a common ratio r between consecutive terms of the sequence, that is

r = 18 ÷ 2 = 162 ÷ 18 = 1458 ÷ 162 = 13122 ÷ 1458 = 9

A recursive formula allows a term in the sequence to be found by multiplying the previous term by the common ratio, that is

$a_{n}$ = 9 $a_{n-1}$ with a₁ = 2

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\\\\
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\quad 
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$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
~~
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$16010.32\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\
t=years\to &9
\end{cases}
\\\\\\
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\\\\\\
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