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.(Thinking)

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)

Re: Finding the common ratio of G.P.

Thank you.

I'll try that.

(Sun)

Re: Finding the common ratio of G.P.

Thank you again.

That worked, I got r = 6.

(Happy)

Re: Finding the common ratio of G.P.

could u pls show me how u arrive at 6 .

Re: Finding the common ratio of G.P.

Hello, labnof!

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

Quote:

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: .$\displaystyle \begin{Bmatrix}a_3 &=& a + 2d \\ a_5 &=& a + 4d \\ a_{17} &=& a + 16d \end{Bmatrix}$ . Assume $\displaystyle d \ne 0.$

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

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

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

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

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

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

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

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

Re: Finding the common ratio of G.P.

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 .

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