# Question #90e7d

Sep 22, 2016

${a}_{n} = 4 \cdot {4}^{n - 1} , n = 1 , 2 , 3 , \ldots$

or
${a}_{n} = 64 \cdot {\left(\frac{1}{4}\right)}^{n - 1} , n = 1 , 2 , 3 , \ldots$

#### Explanation:

As this is posted under geometric sequences, we will proceed under the assumption that the sequences requested must be geometric. A geometric sequence is a sequence whose ${n}^{\text{th}}$ term is given by $a {r}^{n - 1}$ where $a$ is an initial value and $r$ is a common ratio between terms.

By the above, the first $3$ terms are $a , a r , a {r}^{2}$, and we are given the conditions that
$\left\{\begin{matrix}a r = 16 \\ a + a r + a {r}^{2} = 84\end{matrix}\right.$

Substituting the first condition into the second and subtracting $16$, we can rewrite the second condition as

$a + a {r}^{2} = 68$ or $a \left({r}^{2} + 1\right) = 68$

Dividing this by $a r = 16$, we get

$\frac{a \left({r}^{2} + 1\right)}{a r} = \frac{68}{16}$

$\implies \frac{{r}^{2} + 1}{r} = \frac{17}{4}$

$\implies {r}^{2} - \frac{17}{4} r + 1 = 0$

Applying the quadratic formula gives us

$r = \frac{\frac{17}{4} \pm \sqrt{{\left(\frac{17}{4}\right)}^{2} - 4}}{2} = \frac{17 \pm 15}{8}$

So $r = 4$ or $r = \frac{1}{4}$

From $a r = 16$, we also have $a = \frac{16}{r}$. Thus, by selecting the different possibilities for $r$, we get

$\left(a , r\right) = \left(4 , 4\right)$ or $\left(a , r\right) = \left(64 , \frac{1}{4}\right)$

Checking our answers, our first $3$ terms for $\left(a , r\right) = \left(4 , 4\right)$ would be
$4 , 16 , 64$

and our first $3$ terms for $\left(a , r\right) = \left(64 , \frac{1}{4}\right)$ would be
$64 , 16 , 4$

Note that in each case, the sum of the terms is $84$, as desired.

As a side note, if we do not require a geometric sequence, any sequence ${a}_{1} , {a}_{2} , {a}_{3} , \ldots$ would suffice so long as the second and third terms satisfy ${a}_{2} = 16$ and ${a}_{3} = 68 - {a}_{1}$.