Question

Please help? Fan and medal will be given.

When radioactive substances decay, the amount remaining will form a geometric sequence when measured over constant intervals of time. The table shows the amount of a radioactive isotope initially and after 2 hours. What are the amounts left after 1 hour, 3 hours, and 4 hours?

Hours Elapsed - 0, 1, 2, 3, 4
Grams of the isotope - 1986, ___, 477, ___, ___

How much of the isotope is left after 1 hour?
How many grams?

Thank you so much in advance!

Answer 1

Is -1986 is amount at initial time?

Answer 2

Sorry, not negative, it's positive. I just put a slash there to indicate how many hours were elapsed.

Answer 3

Here a1= 1986, a3=477 and r=?. Right

Answer 4

I think so. Yes :) and I apologize for the late reply.

Answer 5

To find "r" use the formula an=a1*r^n-1

Answer 6

Okay. So, 1986*r^n-1

Answer 7

Here we have a1=1986 and a3=477
we start with a3=a1*r^3-1
477=(1986)*r^2
477/1986=r^2
0.24=r^2
\[\sqrt{0.24}\]=r
r=0.48

Answer 8

Now to find a2
Use a2=a1*r^2-1

Answer 9

1986 *0.48^2-1 = 953.28?

Answer 10

Here are the terms: 1986, 953, 477, 219, 105.

How do I find out how much of the isotope is left after 1 hour and how many grams are left?

Answer 11

you got the 953 somehow right? not sure how you got it, but you got it

Answer 12

Yes :) I got it by multiplying a1 by the common ratio. 1986 * 0.48

Answer 13

that is not what i get

Answer 14

let me try again, maybe i made a mistake

Answer 15

\[1986r^2=477\\
r^2=\frac{477}{1986}\\
r=\sqrt{\frac{477}{1986}}\]

Answer 16

953.28 is the actual answer I get when multiplying the two.

Answer 17

which is close to \(0.48\) but i get \(0.49\)

Answer 18

anyway you have answered the question
assuming these numbers are right (i didn't check them) you have \[ 1986, 953, 477, 219, 105. \]

Answer 19

Yes

Answer 20

first is at 0 hour, second is one hour, third is two hour, etc

Answer 21

Okay