What kind of sequence is that?

Suppose the common difference of $a_{k}$ is $d$ and the common ratio of $b_{k}$ is $r$ . Note that $r \neq 0, 1$ . Also let's denote $a_{1}$ by $a$ and $b_{1}$ by $b$ . Note that $a, b \neq 0$ . We will write $c_{1}, c_{2}, c_{3}$ and $c_{5}$ in terms of $a, b, d$ and $r$ .

$\begin{aligned} c_{1} & = a \cdot b = 1 \\ c_{2} & = (a + d) \cdot br = 2 \\ c_{3} & = (a + 2d) \cdot b r^{2} = 3 \\ c_{5} & = (a + 4d) \cdot b r^{4} = \ ? \end{aligned}$

Let's use the $c_{1}$ equation to eliminate $b$ . $\left( b = \frac{1}{a} \right)$ .

$\begin{aligned} c_{2} & = \left(1 + \frac{d}{a} \right) \cdot r = 2 \\ c_{3} & = \left(1 + \frac{2d}{a} \right) \cdot r^{2} = 3 \\ c_{5} & = \left(1 + \frac{4d}{a} \right) \cdot r^{4} = \ ? \end{aligned}$

From the $c_{2}$ equation, we can use the relation $\frac{d}{a} = \frac{2}{r} -1$ in the $c_{3}$ equation.

$c_{3} = \left(1 + \frac{4}{r} -2 \right) \cdot r^{2} = 3$

We get a quadratic equation in $r$ .

$-r^2 + 4r = 3 \implies r ^2 - 4r + 3 = 0 \implies (r -1)(r - 3) = 0$

We know that $r \neq 1$ since $b_{k}$ is a non-constant geometric progression. Hence $r =3$ . We also get $\frac{d}{a}= \frac{2}{3} - 1 = - \frac{1}{3}$ . We substitute these values in the equation of $c_{5}$ :

$c_{5} = \left(1 + \frac{4 \times (-1)}{3} \right) \cdot 3^{4} = \frac{-1}{3} \times 3^4 = \boxed{-27} \ \ _\square$

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Just for fun, we shall attempt to find out the general term of $c_{k}$ .

We have obtained that $a = 3$ and $d = -1$ . Thus, using the formula for arithmetic progressions $a_n = a + (n - 1) d$ , we can say that the arithmetic progression $a_{k} = 3 + (k - 1) (1) = 4 - k$

Similarly, we have also obtained $A = \frac{1}{3}$ , $r = 3$ . Geometric progressions can be represented as $a_{n} = a \times r^{n -1 }$ . Thus the geometric progression $b_{k} = \frac{1}{3} \times 3^{k - 1} = 3^{k - 2}$ .

We can write the general term $c_{k}$ as $a_{k} \cdot b_{k} = (4 - k) \cdot 3^{k - 2}$