0
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ OMG! Thanks a bunch for your help. I really appreciate it. $\endgroup$ – Justin Aug 12 at 7:34
1
$\begingroup$

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 $q=\frac{10}{9}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.