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 . [
Accepted Solution
A:
To solve the first part we are going to use the formula for the nth therm of geometric sequence: [tex]a_{n}=ar^{n-1}[/tex] where [tex]a_{n}[/tex] is the nth term [tex]a_{1}[/tex] is the first term [tex]r[/tex] is the ratio [tex]n[/tex] is the position of the term in the sequence
a. The ratio of a geometric sequence is [tex]r= \frac{a_{n}}{a_{n-1}} [/tex]. We know for our problem that [tex]a_n=u_{8}=36[/tex] and [tex]a_{n-1}=u_{7}=108[/tex]. Lets replace those values in our formula to find [tex]r[/tex]: [tex]r= \frac{36}{108} [/tex] [tex]r= \frac{1}{3} [/tex]
We can conclude that the ratio of our geometric sequence is [tex]r= \frac{1}{3} [/tex].
b. To find [tex]a_{1}[/tex] we are going to use the formula for the nth therm of geometric sequence, the ratio, and the given fact that [tex]u_{7}=108[/tex]: [tex]a_{n}=ar^{n-1}[/tex] [tex]108=a_{1}( \frac{1}{3})^{7-1} [/tex] [tex]108=a_{1}( \frac{1}{3})^{6} [/tex] [tex]108=a_{1}( \frac{1}{729} )[/tex] [tex]a_{1}= \frac{108}{ \frac{1}{729} } [/tex] [tex]a_{1}=78732[/tex]
We can conclude that the first therm, [tex]a_{1}[/tex], 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: [tex]S_{k}=a_{1}( \frac{1-r ^k)}{1-r} )[/tex] where [tex]S_{k}[/tex] is the sum of the first [tex]k[/tex] terms [tex]a_{1}[/tex] is the first term [tex]r[/tex] is the common ratio [tex]k[/tex] is the number of terms
We know for our problem that [tex]S_{k}=118096[/tex], and we also know for previous calculations that [tex]a_{1}=78732[/tex] and [tex]r= \frac{1}{3} [/tex]. So lets replace those values in our formula to find [tex]k[/tex]: [tex]S_{k}=a_{1}( \frac{1-r ^k)}{1-r} )[/tex] [tex]118096=78732[ \frac{1-( \frac{1}{3})^k }{1- \frac{1}{3} } ][/tex] [tex] \frac{118096}{78732} = \frac{1-( \frac{1}{3})^k }{ \frac{2}{3} } [/tex] [tex]k=10[/tex]
We can conclude the the sum of the first 10 terms of our geometric sequence is 118096.