MATH SOLVE

3 months ago

Q:
# Find a1, for the given geometric series. Round to the nearest hundredth if necessary. Sn= 44,240, r= 3.8, n= 9

Accepted Solution

A:

[tex]\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\ \cline{1-1} n=9\\ r=3.8\\ \stackrel{S_n}{S_9}=44240 \end{cases}[/tex][tex]\bf 44240=a_1\left( \cfrac{1-3.8^9}{1-3.8} \right)\implies 44240\approx a_1\left( \cfrac{-165215.101}{-2.8} \right) \\\\\\ 44240\approx a_1(59005.393)\implies \cfrac{44240}{59005.393}\approx a_1\implies \stackrel{\textit{rounded up}}{0.75=a_1}[/tex]