# The fourth, seventh and sixteenth terms of an arithmetic sequence also form consecutive terms of a geometric sequence.

Find the common ratio of the geometric sequence.

Now how do you display the consecutive terms of a geometric sequence as you don't know what the power of $$r$$ is, or do you use like $$r^{(n+1)}$$ , $$r^{(n+2)}$$, so on...

• what is exactly your geometric series?
– user715522
Jan 10, 2020 at 10:39
• if your series is $r^{(n+1)}, r^{(n+2)}$... then the common ratio is $r$
– user715522
Jan 10, 2020 at 10:41
• That is the series Jan 10, 2020 at 10:41
• A geometric series is in the from $a,aq,aq^2,...$ where $a$ is the first term of the geometric series and $q$ is the common ratio, you can find the common ratio by dividing two consecutive terms ,e.g.$\frac{aq^{n}}{aq^{\left(n-1\right)}}$ where $aq^{n}$ represents the $n$th term of the series, here do the same thing and get the common ratio.
– user715522
Jan 10, 2020 at 10:44
• Are you sure you've copied the question correctly? Solving this problem comes down to solving the equation $x^4 - 2x + 1$, and besides the trivial solution $x = 1$ the other solution is rather... ugly. Jan 10, 2020 at 10:46

If the arithmetic sequence has first term $$a$$ and common difference $$d$$ then its fourth term is $$a+3d$$, its seventh term is $$a+6d$$ and its sixteenth term is $$a+15d$$. If these three terms form a geometric sequence with common ratio $$r$$ then

$$r = \frac{a+6d}{a+3d}=\frac{a+15d}{a+6d} \\ \Rightarrow (a+6d)^2 = (a+3d)(a+15d) \\ \Rightarrow a^2 +12ad + 36d^2 = a^2 +18ad + 45d^2$$

I'll let you take it from there.

• Thanks man really helpful. 've just realized Jan 10, 2020 at 11:09

$$a_4,a_7,a_{16}$$ are in GP. $$\implies a_7^2=a_4a_{16}$$ $$\implies (a+6d)^2=(a+3d)(a+15d)$$ Let $$\dfrac{a}{d}=m$$, then $$m^2+12m+36= m^2+18m+45$$ $$\implies m=\dfrac{-3}{2}$$ $$\implies \text{Common ratio = }\dfrac{a_7}{a_4}=\dfrac{a+6d}{a+3d}$$ $$=\dfrac{m+6}{m+3}$$ $$=3$$

• What if $d=0$ ? Jan 10, 2020 at 11:02
• Why is a7 squared equals to a4a16 if i may ask Jan 10, 2020 at 11:03
• and what is a/d=m ? Jan 10, 2020 at 11:04
• @gandalf61, if $d=0$, it is a constant sequence. Hence the common ratio is $1$. But that is a trivial case. Jan 10, 2020 at 11:33
• @Exodus, For $a_4,a_7, a_{16}$ to be in GP, ratio of consecutive terms should be same. Hence, $\dfrac{a_7}{a_4}=\dfrac{a_{16}}{a_7}\implies a_7^2=a_4a_{16}.$ Jan 10, 2020 at 11:34

Edit: I misread the problem. This post solves the question if you exchange the words "arithmetic" and "geometric" in the question. It is much harder than the question that was actually asked.

Indeed. The sequence goes $$a, ar, \ldots, ar^4, \ldots, ar^7, \ldots, ar^{16}, \ldots$$. Now we have to find $$r$$. The assumption is that $$ar^7 - ar^4 = ar^{16} - ar^{7}.$$ Dividing both sides of this equation by $$ar^4$$, and rearranging some, we get $$r^{12} - 2r^3 + 1 = 0.$$ Note that we have the trivial solution $$r = 1$$, corresponding to the constant sequence $$a, a, a, \ldots$$. So let's keep that one in mind. (In fact, there are also trivial solutions when $$a = 0$$ or $$r = 0$$, which we discarded when dividing by $$ar^4$$.) However, there is another solution. Write $$x = r^3$$ to simplify our equation to the quartic $$x^4 - 2x + 1 = 0,$$ and we know that quartic equations have roots in terms of radicals -- but I'd rather not do it by hand, so let's let WolframAlpha help: the real solution, other than $$x = 1$$, is $$x = \frac13 \left(-1 - \frac{2}{\sqrt{17 + 3 \sqrt{33}}} + \sqrt{17 + 3 \sqrt{33}}\right),$$ so that the nontrivial solution gives $$r = \sqrt{\frac13 \left(-1 - \frac{2}{\sqrt{17 + 3 \sqrt{33}}} + \sqrt{17 + 3 \sqrt{33}}\right)}.$$

• This is backwards. The problem says that three terms of an AP are in geometric proportion, and you are finding three terms of a GP that are in arithmetic proportion.
– user694818
Jan 10, 2020 at 11:09
• That explains why the solution was so ugly. :-) I'll put a disclaimer at the top, but might as well leave this up. Jan 10, 2020 at 11:12
• I don't blame you!
– user694818
Jan 10, 2020 at 11:13

The 4th, 7th and 16th terms of any AP can be expressed as $$x, x+y$$ and $$x+4y$$.

We then require $$(x+y)^2=x(x+4y)$$ i.e. $$y=0$$ or $$2x$$.

Ignoring the trivial solution we then have common ratio $$\frac{3x}{x}=3.$$