Question

A radioactive substance decreases in the amount of grams by one-third each year. If the starting amount of the

Answer 1

Answer:

The sequence is geometric. The recursive formula is $a_{n}=2/3a_{n-1}$

Step-by-step explanation:

In order to solve this problem, you have to calculate the amount of the substance left after the end of each year to obtain a sequence and then you have to determine if the sequence is arithmetic or geometric.

The substance decreases by one-third each year, therefore:

After 1 year:

$1452-\frac{1}{3}(1452)$

Using 1452 as a common factor and solving the fraction:

$1452(1-\frac{1}{3})=1452(\frac{2}{3})=968$

You can notice that in general, after each year the amount of grams is the initial amount of the year multiplied by 2/3

After 2 years:

$968(\frac{2}{3})=\frac{1936}{3}$

After 3 years:

$\frac{1936}{3}(\frac{2}{3})=\frac{3872}{9}$

The sequence is:

1452,968,1936/3,3872/9....

In order to determine if the sequence is geometric, you have to calculate the ratio of two consecutive terms and see if the ratio is the same for all two consecutive terms. The ratio is obtained by dividing a term by the previous term.

The sequence is arithmetic if the difference of two consecutive terms is the same for all two consecutive terms.

-Calculating the ratio:

For the first and second terms:

968/1452=2/3

For the second and third terms:

1936/3 ÷ 968 = 2/3

In conclussion, the sequence is geometric because the ratio is common.

The recursive formula of a geometric sequence is given by:

$a_{n}=ra_{n-1}$

where an is the nth term, r is the common ratio and an-1 is the previous term.

In this case, r=2/3