# Math Help - Finding the common ratio of G.P.

1. ## Finding the common ratio of G.P.

Hello,

I am struggling with the following question: 'The third, fifth and seventeenth terms of an A.P. are in geometric progression. Find the common ratio of the G.P.

I have done the following:

Let the third, fifth and seventeenth terms be x, y and z respectively. From which I have, from A.P:

x + 2d = y

x + 14d = z

y + 12d = z

Also from G.P:

xr = y

xr^2 = z

yr = z

xz = y^2

I have three unknowns and seven equations, which I think should be enough, but I seem to be going round in circles. I would very much appreciate it if someone could point me in the right direction and/or suggest another approach.

Thank you.

2. ## Re: Finding the common ratio of G.P.

Let Leat AP have 1st term a and common difference d
Then we have (a+4d)/(a+2d)=(a+16d)/(a+4d)
Hence get a in terms of d and sub back in (a+4d)/(a+2d)

3. ## Re: Finding the common ratio of G.P.

Thank you.

I'll try that.

4. ## Re: Finding the common ratio of G.P.

Thank you again.

That worked, I got r = 6.

5. ## Re: Finding the common ratio of G.P.

could u pls show me how u arrive at 6 .

6. ## Re: Finding the common ratio of G.P.

Hello, labnof!

Here is what I did . . . There may be many shorter methods.

The third, fifth and seventeenth terms of an A.P. are in geometric progression.
Find the common ratio of the G.P.

We have these three A.P. terms: . $\begin{Bmatrix}a_3 &=& a + 2d \\ a_5 &=& a + 4d \\ a_{17} &=& a + 16d \end{Bmatrix}$ . Assume $d \ne 0.$

Since they are in G.P., we have: . $\begin{Bmatrix}r &=& \dfrac{a+4d}{a+2d} & [1] \\ \\[-3mm] r &=& \dfrac{a+16d}{a+4d} & [2] \end{Bmatrix}$

Equate [1] and [2]: . $\frac{a+4d}{a+2d} \:=\:\frac{a+16d}{a+4d} \quad\Rightarrow\quad (a+4d)^2 \:=\:(a+2d)(a+16d)$

. . . . . . . . . . . . . . . $a^2 + 8ad + 16d^2 \:=\:a^2 + 18ad + 32d^2$

. . . . . . . . . . . . . . . $10ad + 16d^2 \:=\:0 \quad\Rightarrow\quad 2d(5a + 8d) \:=\:0$

We have: . $5a + 8d \:=\:0 \quad\Rightarrow\quad d \:=\:\text{-}\tfrac{5}{8}a$

Hence: . $\begin{Bmatrix}a_3 &=& a + 2d &=& a + 2(\text{-}\frac{5}{8}a) &=& \text{-}\frac{1}{4}a \\ \\[-3mm] a_5 &=& a + 4d &=& a + 4(\text{-}\frac{5}{8}a) &=& \text{-}\frac{3}{2}a \\ \\[-3mm] a_{17} &=& a + 16d &=& a + 16(\text{-}\frac{5}{8}a) &=& \text{-}9a \end{Bmatrix}$

Therefore: . $r \;=\;\frac{a_5}{a_3} \;=\;\frac{\text{-}\frac{3}{2}a}{\text{-}\frac{1}{4}a} \quad\Rightarrow\quad \boxed{r \;=\;6}$

Check: . $r \;=\;\frac{a_{17}}{a_5} \;=\;\frac{\text{-}9a}{\text{-}\frac{3}{2}a} \quad\Rightarrow\quad r \:=\:6\;\;\checkmark$

7. ## Re: Finding the common ratio of G.P.

Thank u very much.

8. ## Re: Finding the common ratio of G.P.

i have 2 more problems
1. A geometric series has the first term 2 and common ratio 3/2
Find the greatest number of
terms the series can have without its sum exceeding 125.
2.. A geometric series has the first term 5 and common ratio 3. Find the least number of terms
the series can have if its sum exceeds 2000.

but for the first one i got 8.522 as the answer and 6.085 for the 2nd . but my tutor has a different answer 8 and 7 respectively and i donot know how she came about it could anyone show me .i think she must have approximated but i dont know what she has done precisely .