Me possibly overthinking a geometric series question
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I am creating a study sheet for geometric series and came across this questions
Do you think there is an infinite geometric series with first term 10 and a sum of 4? If so, find one infinite geometric series with first term 10 and a sum of 4, as well as explain how you get this infinite geometric series. If not, explain why.
This is possible with some neagitve common ratio, r, where $|r|<1$ correct?
geometric-series
asked Dec 4 '18 at 0:31
K MathK Math
60039
$endgroup$
add a comment |
$begingroup$
I am creating a study sheet for geometric series and came across this questions
Do you think there is an infinite geometric series with first term 10 and a sum of 4? If so, find one infinite geometric series with first term 10 and a sum of 4, as well as explain how you get this infinite geometric series. If not, explain why.
This is possible with some neagitve common ratio, r, where $|r|<1$ correct?
geometric-series
asked Dec 4 '18 at 0:31
K MathK Math
60039
$endgroup$
$begingroup$
What have you tried so far? What $r$ do you have in mind?
$endgroup$
– platty
Dec 4 '18 at 0:33
add a comment |
$begingroup$
I am creating a study sheet for geometric series and came across this questions
Do you think there is an infinite geometric series with first term 10 and a sum of 4? If so, find one infinite geometric series with first term 10 and a sum of 4, as well as explain how you get this infinite geometric series. If not, explain why.
This is possible with some neagitve common ratio, r, where $|r|<1$ correct?
geometric-series
asked Dec 4 '18 at 0:31
K MathK Math
60039
$endgroup$
I am creating a study sheet for geometric series and came across this questions
Do you think there is an infinite geometric series with first term 10 and a sum of 4? If so, find one infinite geometric series with first term 10 and a sum of 4, as well as explain how you get this infinite geometric series. If not, explain why.
This is possible with some neagitve common ratio, r, where $|r|<1$ correct?
geometric-series
geometric-series
asked Dec 4 '18 at 0:31
K MathK Math
60039
asked Dec 4 '18 at 0:31
K MathK Math
60039
asked Dec 4 '18 at 0:31
K MathK Math
60039
asked Dec 4 '18 at 0:31
K MathK Math
60039
asked Dec 4 '18 at 0:31
K MathK Math
60039
60039
$begingroup$
What have you tried so far? What $r$ do you have in mind?
$endgroup$
– platty
Dec 4 '18 at 0:33
add a comment |
$begingroup$
What have you tried so far? What $r$ do you have in mind?
$endgroup$
– platty
Dec 4 '18 at 0:33
$begingroup$
What have you tried so far? What $r$ do you have in mind?
$endgroup$
– platty
Dec 4 '18 at 0:33
$begingroup$
What have you tried so far? What $r$ do you have in mind?
$endgroup$
– platty
Dec 4 '18 at 0:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You know that if $a$ is the first term of a geometric series and $r$ is the ratio, then the sum is given by
$$S=sum_{n=0}^infty ar^n=frac{a}{1-r}$$
So, if $a=10$ and $S=4$, then
$$frac{10}{1-r}=4implies r=-frac{3}{2}$$
but the series does not converge in this case, so the solution is extraneous and the described situation is not possible.
answered Dec 4 '18 at 0:34
FrpzzdFrpzzd
22.5k840108
$endgroup$
$begingroup$
ok this is what I continued to get when I was solving but wasn't sure if I was missing something.
$endgroup$
– K Math
Dec 4 '18 at 0:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You know that if $a$ is the first term of a geometric series and $r$ is the ratio, then the sum is given by
$$S=sum_{n=0}^infty ar^n=frac{a}{1-r}$$
So, if $a=10$ and $S=4$, then
$$frac{10}{1-r}=4implies r=-frac{3}{2}$$
but the series does not converge in this case, so the solution is extraneous and the described situation is not possible.
answered Dec 4 '18 at 0:34
FrpzzdFrpzzd
22.5k840108
$endgroup$
$begingroup$
ok this is what I continued to get when I was solving but wasn't sure if I was missing something.
$endgroup$
– K Math
Dec 4 '18 at 0:43
add a comment |
$begingroup$
You know that if $a$ is the first term of a geometric series and $r$ is the ratio, then the sum is given by
$$S=sum_{n=0}^infty ar^n=frac{a}{1-r}$$
So, if $a=10$ and $S=4$, then
$$frac{10}{1-r}=4implies r=-frac{3}{2}$$
but the series does not converge in this case, so the solution is extraneous and the described situation is not possible.
answered Dec 4 '18 at 0:34
FrpzzdFrpzzd
22.5k840108
$endgroup$
$begingroup$
ok this is what I continued to get when I was solving but wasn't sure if I was missing something.
$endgroup$
– K Math
Dec 4 '18 at 0:43
add a comment |
$begingroup$
You know that if $a$ is the first term of a geometric series and $r$ is the ratio, then the sum is given by
$$S=sum_{n=0}^infty ar^n=frac{a}{1-r}$$
So, if $a=10$ and $S=4$, then
$$frac{10}{1-r}=4implies r=-frac{3}{2}$$
but the series does not converge in this case, so the solution is extraneous and the described situation is not possible.
answered Dec 4 '18 at 0:34
FrpzzdFrpzzd
22.5k840108
$endgroup$
You know that if $a$ is the first term of a geometric series and $r$ is the ratio, then the sum is given by
$$S=sum_{n=0}^infty ar^n=frac{a}{1-r}$$
So, if $a=10$ and $S=4$, then
$$frac{10}{1-r}=4implies r=-frac{3}{2}$$
but the series does not converge in this case, so the solution is extraneous and the described situation is not possible.
answered Dec 4 '18 at 0:34
FrpzzdFrpzzd
22.5k840108
answered Dec 4 '18 at 0:34
FrpzzdFrpzzd
22.5k840108
answered Dec 4 '18 at 0:34
FrpzzdFrpzzd
22.5k840108
answered Dec 4 '18 at 0:34
FrpzzdFrpzzd
22.5k840108
22.5k840108
$begingroup$
ok this is what I continued to get when I was solving but wasn't sure if I was missing something.
$endgroup$
– K Math
Dec 4 '18 at 0:43
add a comment |
$begingroup$
ok this is what I continued to get when I was solving but wasn't sure if I was missing something.
$endgroup$
– K Math
Dec 4 '18 at 0:43
$begingroup$
ok this is what I continued to get when I was solving but wasn't sure if I was missing something.
$endgroup$
– K Math
Dec 4 '18 at 0:43
$begingroup$
ok this is what I continued to get when I was solving but wasn't sure if I was missing something.
$endgroup$
– K Math
Dec 4 '18 at 0:43
add a comment |
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$begingroup$
What have you tried so far? What $r$ do you have in mind?
$endgroup$
– platty
Dec 4 '18 at 0:33