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.

  • 1
    $\begingroup$ $U_{10}^2=U_{20}.U_{10}$ 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. $\endgroup$ – Jayant Jha Aug 12 at 7:35
  • $\begingroup$ @JayantJha - Yes, it works. Thanks! $\endgroup$ – Justin Aug 12 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 $R=1+9 \frac{d}{a}=\frac{10}{9}.$

  • $\begingroup$ OMG! Thanks a bunch for your help. I really appreciate it. $\endgroup$ – Justin Aug 12 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 $q=\frac{10}{9}$.

  • $\begingroup$ Thanks, Matteo for your help. This was really helpful. $\endgroup$ – Justin Aug 12 at 7:41

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