Question

The seventh term, u7 , of a geometric sequence is 108. the eighth term, u8 , of the sequence is 36 . (a) write down the common ratio of the sequence. [1 mark] (b) find u1 . [2 marks] the sum of the first k terms in the sequence is 118 096 . (c) find the value of k . [

Answer 1

To solve the first part we are going to use the formula for the nth therm of geometric sequence: $a_{n}=ar^{n-1}$
where 
$a_{n}$ is the nth term
$a_{1}$ is the first term
$r$ is the ratio 
$n$ is the position of the term in the sequence

a. The ratio of a geometric sequence is $r= \frac{a_{n}}{a_{n-1}}$. We know for our problem that $a_n=u_{8}=36$ and $a_{n-1}=u_{7}=108$. Lets replace those values in our formula to find $r$:
$r= \frac{36}{108}$
$r= \frac{1}{3}$

We can conclude that the ratio of our geometric sequence is $r= \frac{1}{3}$.

b. To find $a_{1}$ we are going to use the formula for the nth therm of geometric sequence, the ratio, and the given fact that $u_{7}=108$:
$a_{n}=ar^{n-1}$
$108=a_{1}( \frac{1}{3})^{7-1}$
$108=a_{1}( \frac{1}{3})^{6}$
$108=a_{1}( \frac{1}{729} )$
$a_{1}= \frac{108}{ \frac{1}{729} }$
$a_{1}=78732$

We can conclude that the first therm, $a_{1}$, of our geometric sequence is 78732.

c. To solve this one we are going to use the formula for the sum of the first nth terms of a geometric sequence: $S_{k}=a_{1}( \frac{1-r ^k)}{1-r} )$
where
$S_{k}$ is the sum of the first $k$ terms
$a_{1}$ is the first term 
$r$ is the common ratio 
$k$ is the number of terms 

We know for our problem that $S_{k}=118096$, and we also know for previous calculations that $a_{1}=78732$ and $r= \frac{1}{3}$. So lets replace those values in our formula to find $k$:
$S_{k}=a_{1}( \frac{1-r ^k)}{1-r} )$
$118096=78732[ \frac{1-( \frac{1}{3})^k }{1- \frac{1}{3} } ]$
$\frac{118096}{78732} = \frac{1-( \frac{1}{3})^k }{ \frac{2}{3} }$
$k=10$

We can conclude the the sum of the first 10 terms of our geometric sequence is 118096.