Exercise 39 Page 430

We are given two non-consecutive terms of a geometric sequence.

a_{2} = - 72 and a_{6} = - \frac{1}{1 8}

To find the rule for the

n^{th}

term, we first need to find the common ratio

r .

Let's start by recalling the explicit formula of a geometric sequence.

a_{n} = a_{1} r^{n - 1}

We will substitute the given information into this formula to write a system of equations. By solving the system, we will find the values of

r

and

a_{1} .

{a_{2} = a_{1} r^{2 - 1} a_{6} = a_{1} r^{6 - 1} \Rightarrow {- 72 = a_{1} r - \frac{1}{1 8} = a_{1} r^{5} (I) (II)

To solve the system, we will start by dividing Equation (II) by Equation (I). By doing this, we will eliminate

a_{1}

and find the value of

r .

\frac{- \frac{1}{1 8}}{- 7 2} = \frac{a _{1} r ^{5}}{a _{1} r}

▼

Solve for

r

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

\frac{\frac{1}{1 8}}{7 2} = \frac{a _{1} r ^{5}}{a _{1} r}

DivFracD

$\frac{a / c}{b} = \frac{a}{b \cdot c}$

\frac{1}{7 2 ( 1 8 )} = \frac{a _{1} r ^{5}}{a _{1} r}

Multiply

\frac{1}{1 2 9 6} = \frac{a _{1} r ^{5}}{a _{1} r}

ReduceFrac

$\frac{a}{b} = \frac{a / a _{1}}{b / a _{1}}$

\frac{1}{1 2 9 6} = \frac{r ^{5}}{r}

$\frac{a ^{m}}{a} = a^{m - 1}$

\frac{1}{1 2 9 6} = r^{5 - 1}

SubTerm

Subtract term

\frac{1}{1 2 9 6} = r^{4}

RadicalEqn

$4 LHS = 4 RHS$

4 \frac{1}{1 2 9 6} = 4 r^{4}

RootQuot

$4 \frac{a}{b} = \frac{4 a}{4 b}$

\frac{1}{6} = 4 r^{4}

CalcRoot

Calculate root

\frac{1}{6} = \pm r

RearrangeEqn

Rearrange equation

r = \pm \frac{1}{6}

We found out that

r

has two possible solutions,

- \frac{1}{6}

and

\frac{1}{6}

. Knowing the values of

r

, we can substitute them into either of the equations of our system in order to find the value of

a_{1} .

First Solution

Let's substitute

r = - \frac{1}{6}

in Equation (I).

- 72 = a_{1} r

Substitute

$r = - \frac{1}{6}$

- 72 = a_{1} (- \frac{1}{6})

▼

Solve for

a_{1}

MultPosNeg

$a (- b) = - a \cdot b$

- 72 = - a_{1} (\frac{1}{6})

MultEqn

$LHS \cdot 6 = RHS \cdot 6$

- 432 = - a_{1}

AddEqn

$LHS + a_{1} = RHS + a_{1}$

- 432 + a_{1} = 0

AddEqn

$LHS + 432 = RHS + 432$

a_{1} = 432

Finally, we can write the complete formula for the first solution.

a_{n} = a_{1} r^{n - 1} \Rightarrow a_{n} = 432 (- \frac{1}{6})^{n - 1}

Second Solution