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?
Best Answer
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.