Solved on Mar 06, 2024

Radioactive substance decreases by 1/3 each year. Find recursive formula for amount left after $n$ years, where initial amount is $1,452 \mathrm{~g}$ . Is the sequence arithmetic or geometric?

STEP 1

Assumptions
1. The initial amount of the radioactive substance is $1,452 \mathrm{~g}$ .
2. The substance decreases by one-third of its amount each year.
3. A recursive formula is needed to model the amount of substance left after each year.
4. We need to determine whether the sequence formed by the amount of substance left is arithmetic or geometric.

STEP 2

To write a recursive formula, we need to express the amount of substance left after each year in terms of the amount left in the previous year.

STEP 3

Let $a_n$ represent the amount of substance left after $n$ years, and $a_{n-1}$ represent the amount of substance left after $n-1$ years.

STEP 4

Since the substance decreases by one-third each year, we can express $a_n$ as two-thirds of $a_{n-1}$ .
$a_n = \frac{2}{3}a_{n-1}$

STEP 5

We also know the initial amount of the substance, which is the amount left at the start, or after 0 years. We can denote this as $a_0$ .
$a_0 = 1,452 \mathrm{~g}$

STEP 6

The recursive formula will include the initial amount and the rule for finding the amount after each year.
$a_n = \frac{2}{3}a_{n-1}, \text{ for } n \geq 1$ $a_0 = 1,452 \mathrm{~g}$

STEP 7

Now we need to determine whether the sequence is arithmetic or geometric. An arithmetic sequence has a common difference between consecutive terms, while a geometric sequence has a common ratio.

STEP 8

Since the amount of substance decreases by a factor (multiplied by $\frac{2}{3}$ ) each year, this indicates a common ratio, not a common difference.

STEP 9

Therefore, the sequence is geometric because each term is found by multiplying the previous term by the constant ratio $\frac{2}{3}$ .
The recursive formula for the sequence that models the amount of the substance left after the end of each year is: $a_n = \frac{2}{3}a_{n-1}, \text{ for } n \geq 1$ $a_0 = 1,452 \mathrm{~g}$
And the sequence is geometric.

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Radioactive substance decreases by 1/3 each year. Find recursive formula for amount left after $n$ years, where initial amount is $1,452 \mathrm{~g}$ . Is the sequence arithmetic or geometric?

STEP 1

STEP 2

To write a recursive formula, we need to express the amount of substance left after each year in terms of the amount left in the previous year.

STEP 3

Let $a_n$ represent the amount of substance left after $n$ years, and $a_{n-1}$ represent the amount of substance left after $n-1$ years.

STEP 4

Since the substance decreases by one-third each year, we can express $a_n$ as two-thirds of $a_{n-1}$ .
$a_n = \frac{2}{3}a_{n-1}$

STEP 5

We also know the initial amount of the substance, which is the amount left at the start, or after 0 years. We can denote this as $a_0$ .
$a_0 = 1,452 \mathrm{~g}$

STEP 6

The recursive formula will include the initial amount and the rule for finding the amount after each year.
$a_n = \frac{2}{3}a_{n-1}, \text{ for } n \geq 1$ $a_0 = 1,452 \mathrm{~g}$

STEP 7

Now we need to determine whether the sequence is arithmetic or geometric. An arithmetic sequence has a common difference between consecutive terms, while a geometric sequence has a common ratio.

STEP 8

Since the amount of substance decreases by a factor (multiplied by $\frac{2}{3}$ ) each year, this indicates a common ratio, not a common difference.

STEP 9

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Radioactive substance decreases by 1/3 each year. Find recursive formula for amount left after nnn years, where initial amount is 1,452 g1,452 \mathrm{~g}1,452 g. Is the sequence arithmetic or geometric?

STEP 1

STEP 2

To write a recursive formula, we need to express the amount of substance left after each year in terms of the amount left in the previous year.

STEP 3

Let ana_nan​ represent the amount of substance left after nnn years, and an−1a_{n-1}an−1​ represent the amount of substance left after n−1n-1n−1 years.

STEP 4

Since the substance decreases by one-third each year, we can express ana_nan​ as two-thirds of an−1a_{n-1}an−1​.an=23an−1a_n = \frac{2}{3}a_{n-1}an​=32​an−1​

STEP 5

We also know the initial amount of the substance, which is the amount left at the start, or after 0 years. We can denote this as a0a_0a0​.a0=1,452 ga_0 = 1,452 \mathrm{~g}a0​=1,452 g

STEP 6

The recursive formula will include the initial amount and the rule for finding the amount after each year.an=23an−1, for n≥1a_n = \frac{2}{3}a_{n-1}, \text{ for } n \geq 1an​=32​an−1​, for n≥1 a0=1,452 ga_0 = 1,452 \mathrm{~g}a0​=1,452 g

STEP 7

Now we need to determine whether the sequence is arithmetic or geometric. An arithmetic sequence has a common difference between consecutive terms, while a geometric sequence has a common ratio.

STEP 8

Since the amount of substance decreases by a factor (multiplied by 23\frac{2}{3}32​) each year, this indicates a common ratio, not a common difference.

STEP 9

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Radioactive substance decreases by 1/3 each year. Find recursive formula for amount left after $n$ years, where initial amount is $1,452 \mathrm{~g}$ . Is the sequence arithmetic or geometric?

Let $a_n$ represent the amount of substance left after $n$ years, and $a_{n-1}$ represent the amount of substance left after $n-1$ years.

Since the substance decreases by one-third each year, we can express $a_n$ as two-thirds of $a_{n-1}$ .
$a_n = \frac{2}{3}a_{n-1}$

We also know the initial amount of the substance, which is the amount left at the start, or after 0 years. We can denote this as $a_0$ .
$a_0 = 1,452 \mathrm{~g}$

The recursive formula will include the initial amount and the rule for finding the amount after each year.
$a_n = \frac{2}{3}a_{n-1}, \text{ for } n \geq 1$ $a_0 = 1,452 \mathrm{~g}$

Since the amount of substance decreases by a factor (multiplied by $\frac{2}{3}$ ) each year, this indicates a common ratio, not a common difference.