# 005 Sample Final A, Question 22

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Question Consider the following sequence,

${\displaystyle -3,1,-{\frac {1}{3}},{\frac {1}{9}},-{\frac {1}{27}},\cdots }$

a. Determine a formula for ${\displaystyle a_{n}}$, the n-th term of the sequence.
b. Find the sum ${\displaystyle \displaystyle {\sum _{k=1}^{\infty }a_{k}}}$

Foundations
1) What type of series is this?
3) In the formula there are some placeholder variables. What is the value of each placeholder?
1) This series is geometric. The giveaway is there is a number raised to the nth power.
2) The desired formulas are ${\displaystyle a_{n}=a\cdot r^{n-1}}$   and   ${\displaystyle S_{\infty }={\frac {a_{1}}{1-r}}}$
3) ${\displaystyle a_{1}}$ is the first term in the series, which is ${\displaystyle -3}$. The value for r is the ratio between consecutive terms, which is ${\displaystyle {\frac {-1}{3}}}$

Step 1:
The sequence is a geometric sequence. The common ratio is ${\displaystyle r={\frac {-1}{3}}}$.
Step 2:
The formula for the nth term of a geometric series is ${\displaystyle a_{n}=ar^{n-1}}$ where ${\displaystyle a}$ is the first term of the sequence.
So, the formula for this geometric series is ${\displaystyle a_{n}=(-3)\left({\frac {-1}{3}}\right)^{n-1}}$.
Step 3:
For geometric series, ${\displaystyle \displaystyle {\sum _{k=1}^{\infty }a_{k}}={\frac {a}{1-r}}}$ if ${\displaystyle |r|<1}$. Since ${\displaystyle |r|={\frac {1}{3}}}$,
we have ${\displaystyle \displaystyle {\sum _{k=1}^{\infty }a_{k}}={\frac {-3}{1-{\frac {-1}{3}}}}={\frac {-9}{4}}}$.
${\displaystyle a_{n}=(-3)\left({\frac {-1}{3}}\right)^{n-1}}$
${\displaystyle {\frac {-9}{4}}}$