# Identify whether the series is an arithmetic series (AS), geometric series (GS) or not one of the difined types ( NASGS ) and determine the sum (for NASGS, write NO SUM). Write your Solution. a. 1+frac32+frac94+...+frac{81}{16}

Identify whether the series is an arithmetic series (AS), geometric series (GS) or not one of the difined types ( NASGS ) and determine the sum (for NASGS, write NO SUM). Write your Solution.
a. $1+\frac{3}{2}+\frac{9}{4}+...+\frac{81}{16}$
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faldduE
Given series is $1+\frac{3}{2}+\frac{9}{4}+...+\frac{81}{16}$
Here, we have
${a}_{1}=1,{a}_{2}=\frac{3}{2},{a}_{3}=\frac{9}{4}$
$\frac{{a}_{2}}{{a}_{1}}=\frac{\left(\frac{3}{2}\right)}{1}=\frac{3}{2}$
$\frac{{a}_{3}}{{a}_{2}}=\frac{\left(\frac{9}{4}\right)}{\left(\frac{3}{2}\right)}=\frac{3}{2}$
We get that $\frac{{a}_{2}}{{a}_{1}}=\frac{{a}_{3}}{{a}_{2}}$.
Sice, the ratio of terms is same.
Therefore, given series is geometric where common ratio is $r=\frac{3}{2}$
Now, we know that n-th term of geometric series is given as:
${a}_{n}={a}_{1}{r}^{n-1}$
$\frac{81}{16}=\left(1\right)\left(\frac{3}{2}{\right)}^{n-1}$
$\left(\frac{3}{2}{\right)}^{4}=\left(\frac{3}{2}{\right)}^{n-1}$
On comparing both sides, we get
$n-1=4$
$n=5$
Therefore, number of terms in series is 5.
Sum of geomteric series is given as:
${S}_{n}=\frac{{a}_{1}\left({r}^{n}-1\right)}{r-1}$
Therefore, sum of the given geometric series will be:
${S}_{5}=\frac{1\left(\left(\frac{3}{2}{\right)}^{5}-1\right)}{\frac{3}{2}-1}$
$=\frac{\frac{243}{32}-1}{\frac{1}{2}}$
$=\frac{\frac{211}{32}}{\frac{1}{2}}$
$=\frac{211}{16}$
Hence, sum is $=\frac{211}{16}$
Jeffrey Jordon