# Combined arithmetic and geometric series question

Here is a combined arithmetic and geometric series question -

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.

[10 marks]

Here's what I have done so far,

$$\Rightarrow\ U_1 = a = V_1$$

$$\Rightarrow\ U_{10} = a + 9d = V_2$$

$$\Rightarrow\ U_{20} = a + 19d = V_3$$

Well, that's all I can derive from the question. I don't know where to go from here.

• $U_{10}^2=U_{20}.U_{10}$ Try using this. You will get know something about a and d. then using ${U_{10} \over U_1}=r$ you can get the common ratio. Commented Aug 12, 2019 at 7:35
• @JayantJha - Yes, it works. Thanks! Commented Aug 12, 2019 at 7:37

Call the common ratio $$R$$. Hence

$$R= \frac{V_2}{V_1}$$ and $$R= \frac{V_3}{V_2}$$.

This gives

$$(a+9d)^2=a^2+19ad$$, hence $$a=81d.$$

Therefore $$R=1+9 \frac{d}{a}=\frac{10}{9}.$$

• OMG! Thanks a bunch for your help. I really appreciate it. Commented Aug 12, 2019 at 7:34

Starting from your attempt, I have: $$\left\{\begin{matrix} a_1=b_1 \\ a_1+9d=b_1\cdot q \\ a_1+19d=b_1\cdot q^2 \end{matrix}\right.$$

Solving, I obtain:

$$\left\{\begin{matrix} a_1=b_1 \\ b_1(q-1)=9d \\ b_1(q-1)(q+1)=19d \end{matrix}\right.$$

Substituting, I obtain:

$$\left\{\begin{matrix} a_1=b_1 \\ a_1+9d=b_1\cdot q \\ 9d(q+1)=19d \end{matrix}\right.$$

From here I have $$q=\frac{10}{9}$$.

• Thanks, Matteo for your help. This was really helpful. Commented Aug 12, 2019 at 7:41