Chapter 12.3, Problem 84E

To determine

To evaluate: The sequence $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\cdots$ is a geometric sequence if $a_1,a_2,a_3,\cdots$ is a geometric sequence with common ratio r and find the common ratio of the sequence $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\cdots$.

Answer to Problem 84E

The common ration of sequence $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\cdots$ is $\frac{a_1}{a_2}$.

Explanation of Solution

Proof:

Section1:

The geometric sequence is given,

$a_1,a_2,a_3,\cdots$ (1)

The common ratio of this sequence is r.

The common ratio determines the ratio of second and first term of the sequence (1) is equal to the ratio of third term and second term of the sequence (1),

$r=\frac{a_2}{a_1}$

And,

$r=\frac{a_3}{a_2}$

So, the relation between terms is,

$\begin{array}{l}\frac{a_2}{a_1}=\frac{a_3}{a_2}\\ a_2{}^2=a_3\cdot a_1\end{array}$

The above relation defines the property of geometric series.

The sequence is given,

$\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\cdots$ (2)

If the above sequence is geometric sequence then it satisfies the relation $a_2{}^2=a_3\cdot a_1$ of sequence (1)

If the sequence is geometric then the ratio of second and first term of the sequence (2) is equal to the ratio of third term and second term of the sequence (2),

So, above statement gives below relation,

$\begin{array}{l}\frac{\frac{1}{a_2}}{\frac{1}{a_1}}=\frac{\frac{1}{a_3}}{\frac{1}{a_2}}\\ \frac{a_1}{a_2}=\frac{a_2}{a_1}\\ a_2{}^2=a_1\cdot a_3\end{array}$

The above relation satisfies the property of geometric sequence.

Thus, the sequence $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\cdots$ is a geometric sequence.

Section2:

From Section1, the sequence $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\cdots$ is a geometric sequence.

The common ratio of this sequence is the ratio of second term and first term,

$\begin{array}{c}{r}^{\text{'}}=\frac{\frac{1}{a_2}}{\frac{1}{a_1}}\\ =\frac{a_1}{a_2}\end{array}$

Thus the common ratio of sequence is $\frac{a_1}{a_2}$.