Finding the common ratio of G.P.

**Furyan** · July 8th 2012, 12:04 PM

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.

**biffboy** · July 8th 2012, 09:08 PM

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)

**Furyan** · July 9th 2012, 01:03 AM

Thank you.

I'll try that.

**Furyan** · July 9th 2012, 05:24 PM

Thank you again.

That worked, I got r = 6.

**labnof** · September 21st 2012, 03:41 PM

could u pls show me how u arrive at 6 .

**Soroban** · September 21st 2012, 05:17 PM

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

**labnof** · September 22nd 2012, 01:39 AM

Thank u very much.

**labnof** · September 22nd 2012, 01:54 AM

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 .

Thread: Finding the common ratio of G.P.

LinkBack

Thread Tools

Display

Finding the common ratio of G.P.

Re: Finding the common ratio of G.P.

Re: Finding the common ratio of G.P.

Re: Finding the common ratio of G.P.

Re: Finding the common ratio of G.P.

Re: Finding the common ratio of G.P.

Re: Finding the common ratio of G.P.

Re: Finding the common ratio of G.P.

Similar Math Help Forum Discussions

common ratio

common ratio of the seqence

Common ratio problems

[SOLVED] common ratio

Finding common ratio and more?

Search Tags