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 $a$. 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 $a$.
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 27 mins ago
YuiTo Cheng
2,1612937
asked 3 hours ago
lollollollol
181
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.
$endgroup$
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 27 mins ago
YuiTo Cheng
2,1612937
asked 3 hours ago
lollollollol
181
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.
$endgroup$
8
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
3 hours ago
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
3 hours ago
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 27 mins ago
YuiTo Cheng
2,1612937
asked 3 hours ago
lollollollol
181
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 27 mins ago
YuiTo Cheng
2,1612937
asked 3 hours ago
lollollollol
181
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 27 mins ago
YuiTo Cheng
2,1612937
asked 3 hours ago
lollollollol
181
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 27 mins ago
YuiTo Cheng
2,1612937
edited 27 mins ago
YuiTo Cheng
2,1612937
edited 27 mins ago
YuiTo Cheng
2,1612937
2,1612937
asked 3 hours ago
lollollollol
181
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 3 hours ago
lollollollol
181
asked 3 hours ago
lollollollol
181
181
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.
8
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
3 hours ago
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
3 hours ago
add a comment |
8
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
3 hours ago
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
3 hours ago
8
8
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
3 hours ago
1
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
3 hours ago
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
3 hours ago
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 2 hours ago
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
2 hours ago
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
2 hours ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
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 2 hours ago
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
2 hours ago
add a comment |
Your Answer
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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 2 hours ago
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
2 hours ago
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
2 hours ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
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 2 hours ago
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
2 hours ago
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
2 hours ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
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 2 hours ago
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 2 hours ago
Robert LewisRobert Lewis
48.5k23167
answered 2 hours ago
Robert LewisRobert Lewis
48.5k23167
answered 2 hours ago
Robert LewisRobert Lewis
48.5k23167
answered 2 hours ago
Robert LewisRobert Lewis
48.5k23167
48.5k23167
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
2 hours ago
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
2 hours ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
add a comment |
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
2 hours ago
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
2 hours ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
1
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
2 hours ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
2 hours ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
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 2 hours ago
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
2 hours ago
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 2 hours ago
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
2 hours ago
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 2 hours ago
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 2 hours ago
rashrash
595116
answered 2 hours ago
rashrash
595116
answered 2 hours ago
rashrash
595116
answered 2 hours ago
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
2 hours ago
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
2 hours ago
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
2 hours ago
$begingroup$
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
2 hours ago
add a comment |
lollol is a new contributor. Be nice, and check out our Code of Conduct.
lollol is a new contributor. Be nice, and check out our Code of Conduct.
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8
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
3 hours ago
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
3 hours ago