Question

A geometric sequence has allpositive terms. The sum of thefirst two terms is 15 and the sumto infinity is 27.a Find the value of the commonratio.b Hence, find the first term.

Answer 1

a) Common ratio = 2/7

b) First term = 135/7

Explanations:

The formula for finding the sum of a geometric progression is expressed as:

`S_n=(a(r^n-1))/(r-1)`

Since the sum of the first two terms is 15, then

S2 = 15

n = 2

Substitute into the formula:

`\begin{gathered} S_2=\frac{a\mleft(r^2-1\mright)}{r^{}-1} \\ 15=\frac{a(r+1)\cancel{r-1}}{\cancel{r-1}} \\ 15=a(r+1) \end{gathered}`

Also, the sum to infinity of a geometric sequence is expressed as:

`\begin{gathered} S_(\infty)=(a)/(1-r) \\ _{} \end{gathered}`

Substitute the given values into the formula:

`27=(a)/(1-r)`

Solve both expressions simultaneously

`\begin{gathered} 15=a(r+1) \\ 27=(a)/(1-r) \end{gathered}`

Divide both expressions to have:

`(15)/(27)=(1-r)/(r+1)`

Cross multiply and solve for the common ratio "r"

`\begin{gathered} 15(r+1)=27(1-r) \\ 15r+15=27-27r \\ 15r+27r=27-15 \\ 42r=12 \\ r=(12)/(42) \\ r=(2)/(7) \end{gathered}`

Hence the value of the common ratio is 2/7

b) Get the first term of the sequence;

Using the formula:

`\begin{gathered} 27=(a)/(1-r) \\ 27=(a)/(1-(2)/(7)) \\ 27=(a)/(((5)/(7))) \\ a=27*(5)/(7) \\ a=(135)/(7) \\ \end{gathered}`

Hence the first term of the sequence is 135/7