Chapter 12, Problem 13T

A puppy weighs 0.85 lb at birth, and each week he gains 24% in weight. Let a_n be his weight in pounds at the end of his nth week of life.

1. (a) Find a formula for a_n.
2. (b) How much does the puppy weigh when he is 6 weeks old?
3. (c) Is the sequence a₁, a₂, a₃, … arithmetic, geometric, or neither?

To determine

To find: The formula for $a_n$ of the sequence generated.

Answer to Problem 13T

The formula of the sequence generated is $\left(0.85\right)\times {\left(1.24\right)}^n$.

Explanation of Solution

Given:

The weight of puppy at birth is $0.85\text{lb}$, increment in his weight each week is $24\%$ and weight in pound at the $n\text{th}$ week is $a_n$.

Calculation:

As there is an increment of weight every week of $24\%$, the sequence after $n\text{th}$ week is,

$\text{Final}\text{weight}=\left(\text{Initial}\text{weight}\right)\text{×}{\left(\text{1}+\text{\%increase}\right)}^n$ (1)

Substitute $0.85$ for Initial weight, $24\%$ for increase and $a_n$ for final weight in equation (1) to generalize the sequence.

$\begin{array}{c}{a}_n=\left(0.85\right)\times {\left(1+24\%\right)}^n\\ =\left(0.85\right)\times {\left(1.24\right)}^n\end{array}$

Thus, formula of $a_n$ for sequence generated is $\left(0.85\right)\times {\left(1.24\right)}^n$.

To determine

To find: Weight of puppy after 6 weeks.

Answer to Problem 13T

Weight of puppy after 6 weeks is $3.09\text{lb}$.

Explanation of Solution

Given:

The weight of puppy at birth is $0.85\text{lb}$, increment in his weight each week is $24\%$ and weight in pound at the $n\text{th}$ week is $a_n$.

Calculation:

From part (a) the formula of the sequence generated is,

$a_n=\left(0.85\right)\times {\left(1.24\right)}^n$ (1)

Substitute 6 for n in equation (1) for the weight of puppy after 6 weeks,

$\begin{array}{c}{a}_6=\left(0.85\right)\times {\left(1.24\right)}^6\\ =3.09\end{array}$

Thus, the weight of puppy after 6 weeks is $3.09\text{lb}$.

To determine

Whether the sequence $a_1,a_2,a_3\cdots$ is arithmetic, geometric or neither.

Answer to Problem 13T

The sequence $a_1,a_2,a_3\cdots$ is geometric in nature.

Explanation of Solution

The sequence to be in arithmetic should have the common difference between two specific term constant with nature linear, for geometric should have common ratio between two specific term constant with nature geometric.

From part (a) the formulae of the sequence generated is,

$a_n=\left(0.85\right)\times {\left(1.24\right)}^n$ (1)

Substitute 1 for n in equation (1) for first term of sequence.

$\begin{array}{c}{a}_1=\left(0.85\right)\times {\left(1.24\right)}^1\\ =1.054\end{array}$

Substitute 2 for n in equation (1) for second term of sequence.

$\begin{array}{c}{a}_2=\left(0.85\right)\times {\left(1.24\right)}^2\\ =1.30696\end{array}$

Substitute 3 for n in equation (1) for third term of sequence.

$\begin{array}{c}{a}_3=\left(0.85\right)\times {\left(1.24\right)}^3\\ =1.6206\end{array}$

The common ratio for first two terms is,

$\begin{array}{c}{r}_1=\frac{1.30696}{1.054}\\ =1.24\end{array}$

The common ratio for second and third terms is,

$\begin{array}{c}{r}_2=\frac{1.6206}{1.30696}\\ =1.24\end{array}$

From the above result the value of $r_1$ and $r_2$ are equal which satisfies the condition for geometric series as the terms proceed geometrically,

Thus, the sequence $a_1,a_2,a_3\cdots$ is geometric in nature.