Can someone help The Next CEO of Stack OverflowIf $(c_n)_n$ is the sum of geometric and...
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Can someone help
The Next CEO of Stack OverflowIf $(c_n)_n$ is the sum of geometric and arithmetic sequences. How to get the original sequences back?Given common terms (and their position) between an arithmetic and geometric sequences, find the common ratio.Series - calculating the sumFirst term of a series with two zeros and a constant second differenceGiven a sequence find nth termFinding which term in a sequence the last term of a sum corresponds to.Consecutive termsWrite 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 $93$sequence with first difference and second constant ratio in first difference
$begingroup$
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 2 hours ago
Lehs
7,07931664
asked 3 hours ago
lollollollol
201
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 2 hours ago
Lehs
7,07931664
asked 3 hours ago
lollollollol
201
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
2 hours ago
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
2 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 2 hours ago
Lehs
7,07931664
asked 3 hours ago
lollollollol
201
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 2 hours ago
Lehs
7,07931664
asked 3 hours ago
lollollollol
201
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 2 hours ago
Lehs
7,07931664
asked 3 hours ago
lollollollol
201
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 2 hours ago
Lehs
7,07931664
edited 2 hours ago
Lehs
7,07931664
edited 2 hours ago
Lehs
7,07931664
7,07931664
asked 3 hours ago
lollollollol
201
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
201
asked 3 hours ago
lollollollol
201
201
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
2 hours ago
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
2 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
2 hours ago
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
2 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
2 hours ago
$begingroup$
So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
$endgroup$
– Minus One-Twelfth
2 hours ago
1
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
2 hours ago
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$a^n + a^{n + 1} = a^{n + 2}; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered 1 hour ago
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
1 hour 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
1 hour ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
add a comment |
$begingroup$
$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$
answered 1 hour 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
1 hour 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
$begingroup$
$a^n + a^{n + 1} = a^{n + 2}; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered 1 hour ago
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
1 hour 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
1 hour ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
add a comment |
$begingroup$
$a^n + a^{n + 1} = a^{n + 2}; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered 1 hour ago
Robert LewisRobert Lewis
48.5k23167
$endgroup$
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
1 hour 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
1 hour ago
$begingroup$
@RossMillikan: yes, I see your point!
$endgroup$
– Robert Lewis
1 hour ago
add a comment |
$begingroup$
$a^n + a^{n + 1} = a^{n + 2}; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered 1 hour ago
Robert LewisRobert Lewis
48.5k23167
$endgroup$
$a^n + a^{n + 1} = a^{n + 2}; tag 1$
$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$
$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$
note that
$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$
answered 1 hour ago
Robert LewisRobert Lewis
48.5k23167
answered 1 hour ago
Robert LewisRobert Lewis
48.5k23167
answered 1 hour ago
Robert LewisRobert Lewis
48.5k23167
answered 1 hour ago
Robert LewisRobert Lewis
48.5k23167
48.5k23167
1
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
1 hour 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
1 hour 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
1 hour ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
1 hour 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
1 hour 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
1 hour ago
$begingroup$
In fact, $a=0$ is also a solution.
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
1 hour ago
$begingroup$
@RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
$endgroup$
– Robert Lewis
1 hour 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
1 hour 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
1 hour 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 |
$begingroup$
$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$
answered 1 hour 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
1 hour ago
add a comment |
$begingroup$
$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$
answered 1 hour 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
1 hour ago
add a comment |
$begingroup$
$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$
answered 1 hour ago
rashrash
595116
$endgroup$
$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$
answered 1 hour ago
rashrash
595116
answered 1 hour ago
rashrash
595116
answered 1 hour ago
rashrash
595116
answered 1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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.
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
2 hours ago
1
$begingroup$
Son of Bonacci would know the answer right away.
$endgroup$
– dnqxt
2 hours ago