# A combined arithmetic and geometric sequence question

Here is a question I am currently struggling with -

The first, the tenth and the twentieth terms of an increasing arithmetic sequence are also consecutive terms in an increasing geometric sequence. Find the common ratio of the geometric sequence.

Here's what I've done so far -

$$u_1=v_1$$
$$u_{10}=v_2$$
$$u_{20}=v_3$$

We know that,

$$\displaystyle \frac{v_2}{v_1}=\frac{v_3}{v_2}$$

and,

$$u_1=u_1$$
$$u_{10}=u_1+9d$$
$$u_{20}=u_1+19d$$

Therefore,

$$\displaystyle \frac{u_1+9d}{u_1}=\frac{u_1+19d}{u_1+9d}$$

Upon simplifying -

$$(u_1+9d)^{2}=u_1(u_1+19d)$$

$$\therefore \displaystyle u_1=81d$$

Now, what do I do after this?

• There are no series in your question. May 19 at 17:15
• @José: The title is the general topic under which the question comes, and there wasn't a tag for geometric sequence, so I used [geometric-series] instead.. I've removed it now.. May 19 at 17:18
• What do you do after this? Did you try substituting in your value for $u_1$ into your expressions for the ratio? May 19 at 17:19
• @FShrike: I was stuck because I wasn't aware of any other ratios that existed at the time of writing this question. As Josè points out in their answer, the ratio I was supposed to put the value of $u_1$ into was $\frac{u_{10}}{u_1} = \frac{u_1+9d}{u_1}$... May 19 at 17:35

You have then$$\frac{v_2}{v_1}=\frac{u_{10}}{u_1}=\frac{90d}{81d}=\frac{10}9\quad\text{and}\quad\frac{v_3}{v_2}=\frac{u_{20}}{u_{10}}=\frac{100d}{90d}=\frac{10}9.$$Therefore, the answer is $$\frac{10}9$$.

Note that there is really no need to do what comes after “and”. It's just to double check things.

We could also set up the relations between the terms of the geometric sequence in this way:

$$v_2 \ \ = \ \ r·u_1 \ \ = \ \ u_1 \ + \ 9·d \ \ \ , \ \ \ v_3 \ \ = \ \ r^2·u_1 \ \ = \ \ u_1 \ + \ 19·d$$ $$\Rightarrow \ \ (r \ - \ 1)·u_1 \ \ = \ \ 9·d \ \ \ , \ \ \ (r^2 \ - \ 1)·u_1 \ \ = \ \ 19·d$$ $$\Rightarrow \ \ \frac{(r^2 \ - \ 1)·u_1}{(r \ - \ 1)·u_1} \ \ = \ \ r \ + \ 1 \ \ = \ \ \frac{19·d}{9·d} \ \ \Rightarrow \ \ r \ \ = \ \ \frac{19}{9} \ - \ 1 \ \ = \ \ \frac{10}{9} \ \ .$$

This is not the description of a unique sequence: the problem statement applies equally well to

$$\ \ \ d \ = \ \frac19 \ \ \rightarrow \ \ u_1 \ = \ 9 \ \ , \ \ u_{10} \ = \ 10 \ \ , \ \ u_{20} \ = \ \frac{100}{9} \ \ \$$ or to

$$\ \ \ d \ = \ -3 \ \ \rightarrow \ \ u_1 \ = \ -243 \ \ , \ \ u_{10} \ = \ -270 \ \ , \ \ u_{20} \ = \ -300 \ \ .$$

In this approach, $$\ u_1 \$$ and $$\ d \$$ are "eliminated in one stroke", which is not important here since we are not asked to determine them. (The fact that this happens is the "cause" of the sequence not being unique.) Once we know $$\ r \ \ ,$$ we can find the relation between these two quantities by inserting it into, for instance, $$\ \left(\frac{10}{9} \ - \ 1 \right)·u_1 \ \ = \ \ 9·d \ \Rightarrow \ u_1 \ = \ 81·d \ \ , \$$ as you observed.