Suppose that the sum of two consecutive terms in the the geometric sequence $1, a, a^2,dots$ gives the next term in the sequence. Find $a$. [on hold] The Next CEO of Stack OverflowFind the 12th term and the sum of the first 12 terms of a geometric sequence.Prove consecutive terms in a geometric sequence and consecutive terms in an arithmetic sequence.Given common terms (and their position) between an arithmetic and geometric sequences, find the common ratio.The sum of the first $3$ terms is $24$ and the sum of the next $3$ terms is $ 51.$Find the sum of the geometric sequenceFinding the next term in a sequence.Finding which term in a sequence the last term of a sum corresponds to.Show that the common ratio between any two consecutive terms of a geometric sequence is a constant r.Write the first ten terms of the arithmetic sequence given the first term and some other informationFind three numbers that can be consecutive terms of geometric sequence and first, second and seventh term of arithmetic sequence and whose sum is $93$
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Suppose that the sum of two consecutive terms in the the geometric sequence $1, a, a^2,dots$ gives the next term in the sequence. Find $a$. [on hold]
The Next CEO of Stack OverflowFind the 12th term and the sum of the first 12 terms of a geometric sequence.Prove consecutive terms in a geometric sequence and consecutive terms in an arithmetic sequence.Given common terms (and their position) between an arithmetic and geometric sequences, find the common ratio.The sum of the first $3$ terms is $24$ and the sum of the next $3$ terms is $ 51.$Find the sum of the geometric sequenceFinding the next term in a sequence.Finding which term in a sequence the last term of a sum corresponds to.Show that the common ratio between any two consecutive terms of a geometric sequence is a constant r.Write the first ten terms of the arithmetic sequence given the first term and some other informationFind three numbers that can be consecutive terms of geometric sequence and first, second and seventh term of arithmetic sequence and whose sum is $93$
$begingroup$
Consider the geometric sequence $1, a, a^2, a^3,dots$ Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find $a$.
sequences-and-series
edited yesterday
YuiTo Cheng
2,1862937
asked yesterday
lollollollol
131
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$endgroup$
put on hold as off-topic by Moishe Kohan, Saad, John Omielan, Martin R, Jyrki Lahtonen yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Moishe Kohan, Saad, John Omielan, Martin R, Jyrki Lahtonen
add a comment |
$begingroup$
Consider the geometric sequence $1, a, a^2, a^3,dots$ Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find $a$.
sequences-and-series
edited yesterday
YuiTo Cheng
2,1862937
asked yesterday
lollollollol
131
New contributor
lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Moishe Kohan, Saad, John Omielan, Martin R, Jyrki Lahtonen yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Moishe Kohan, Saad, John Omielan, Martin R, Jyrki Lahtonen
9
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
$begingroup$
Consider the geometric sequence $1, a, a^2, a^3,dots$ Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find $a$.
sequences-and-series
edited yesterday
YuiTo Cheng
2,1862937
asked yesterday
lollollollol
131
New contributor
lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Consider the geometric sequence $1, a, a^2, a^3,dots$ Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find $a$.
sequences-and-series
sequences-and-series
edited yesterday
YuiTo Cheng
2,1862937
asked yesterday
lollollollol
131
New contributor
lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
YuiTo Cheng
2,1862937
asked yesterday
lollollollol
131
New contributor
lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
YuiTo Cheng
2,1862937
edited yesterday
YuiTo Cheng
2,1862937
edited yesterday
YuiTo Cheng
2,1862937
2,1862937
asked yesterday
lollollollol
131
New contributor
lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
lollollollol
131
asked yesterday
lollollollol
131
131
New contributor
lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Moishe Kohan, Saad, John Omielan, Martin R, Jyrki Lahtonen yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Moishe Kohan, Saad, John Omielan, Martin R, Jyrki Lahtonen
put on hold as off-topic by Moishe Kohan, Saad, John Omielan, Martin R, Jyrki Lahtonen yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Moishe Kohan, Saad, John Omielan, Martin R, Jyrki Lahtonen
9
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
9
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
yesterday
9
9
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$a^n + a^n + 1 = a^n + 2; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered yesterday
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
yesterday
1
$begingroup$
I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
yesterday
add a comment |
$begingroup$
$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.
Hence $a = frac 1 pm sqrt52$
answered yesterday
rashrash
595116
$endgroup$
1
$begingroup$
The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
$endgroup$
– Ross Millikan
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$a^n + a^n + 1 = a^n + 2; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered yesterday
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
yesterday
1
$begingroup$
I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
yesterday
add a comment |
$begingroup$
$a^n + a^n + 1 = a^n + 2; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered yesterday
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
yesterday
1
$begingroup$
I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
yesterday
add a comment |
$begingroup$
$a^n + a^n + 1 = a^n + 2; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered yesterday
Robert LewisRobert Lewis
48.5k23167
$endgroup$
$a^n + a^n + 1 = a^n + 2; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered yesterday
Robert LewisRobert Lewis
48.5k23167
answered yesterday
Robert LewisRobert Lewis
48.5k23167
answered yesterday
Robert LewisRobert Lewis
48.5k23167
answered yesterday
Robert LewisRobert Lewis
48.5k23167
48.5k23167
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
yesterday
1
$begingroup$
I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
yesterday
add a comment |
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
yesterday
1
$begingroup$
I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
yesterday
1
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
yesterday
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
yesterday
1
1
$begingroup$
I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
yesterday
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
yesterday
add a comment |
$begingroup$
$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.
Hence $a = frac 1 pm sqrt52$
answered yesterday
rashrash
595116
$endgroup$
1
$begingroup$
The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
$endgroup$
– Ross Millikan
yesterday
add a comment |
$begingroup$
$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.
Hence $a = frac 1 pm sqrt52$
answered yesterday
rashrash
595116
$endgroup$
1
$begingroup$
The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
$endgroup$
– Ross Millikan
yesterday
add a comment |
$begingroup$
$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.
Hence $a = frac 1 pm sqrt52$
answered yesterday
rashrash
595116
$endgroup$
$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.
Hence $a = frac 1 pm sqrt52$
answered yesterday
rashrash
595116
answered yesterday
rashrash
595116
answered yesterday
rashrash
595116
answered yesterday
rashrash
595116
595116
1
$begingroup$
The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
$endgroup$
– Ross Millikan
yesterday
add a comment |
1
$begingroup$
The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
$endgroup$
– Ross Millikan
yesterday
1
1
$begingroup$
The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
$endgroup$
– Ross Millikan
yesterday
$begingroup$
The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
$endgroup$
– Ross Millikan
yesterday
add a comment |
9
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
yesterday