Question

The sum of the first 10 terms of a geometric sequence is 4−2^−8. If the common ratio is 1/2, find the first term of the sequence. Use this to determine the 11th term of the sequence.

Answer 1

`\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}\\[-0.5em] \hrulefill\\ ~~ r=(1)/(2)\\ ~~ n=10\\ S_(10)=4-(1)/(2^8) \end{cases}`

`S_(10)=a_1\left( \cfrac{1-\left( (1)/(2) \right)^(10)}{1-\left( (1)/(2) \right)} \right)\implies 4-\cfrac{1}{2^8}=a_1\left( \cfrac{1-\left( (1)/(2) \right)^(10)}{1-\left( (1)/(2) \right)} \right) \\\\\\ 4-\cfrac{1}{2^8}=a_1\left( \cfrac{ ~~ (1023)/(1024) ~~ }{(1)/(2)} \right)\implies \cfrac{1023}{256}~~ = ~~a_1\left( \cfrac{1023}{512} \right) \\\\\\ \left( \cfrac{512}{1023} \right)\cfrac{1023}{256}~~ = ~~a_1\implies \boxed{2=a_1} \\\\[-0.35em] ~\dotfill`

`n^(th)\textit{ term of a geometric sequence} \\\\ a_n=a_1\cdot r^(n-1)\qquad \begin{cases} a_n=n^(th)\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\[-0.5em] \hrulefill\\ a_1=2\\ n=11\\ r=(1)/(2) \end{cases} \\\\\\ a_(11)=2\left( (1)/(2) \right)^(11-1)\implies a_(11)=2\left( (1)/(2) \right)^(10)\implies a_(11)=\cfrac{2}{2^(10)}\implies \boxed{a_(11)=\cfrac{1}{512}}`