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












2












$begingroup$


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







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New contributor




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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
















2












$begingroup$


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







share|cite|improve this question









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














2












2








2


2



$begingroup$


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







share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Lehs

7,07931664




7,07931664






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 2 hours ago









lollollollol

221




221




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














  • 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










2 Answers
2






active

oldest

votes


















3












$begingroup$

$a^n + a^{n + 1} = a^{n + 2}; tag 1$



$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$



$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$



note that



$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$






share|cite|improve this answer









$endgroup$









  • 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



















3












$begingroup$

$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$






share|cite|improve this answer









$endgroup$









  • 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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

$a^n + a^{n + 1} = a^{n + 2}; tag 1$



$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$



$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$



note that



$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$






share|cite|improve this answer









$endgroup$









  • 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
















3












$begingroup$

$a^n + a^{n + 1} = a^{n + 2}; tag 1$



$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$



$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$



note that



$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$






share|cite|improve this answer









$endgroup$









  • 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














3












3








3





$begingroup$

$a^n + a^{n + 1} = a^{n + 2}; tag 1$



$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$



$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$



note that



$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$






share|cite|improve this answer









$endgroup$



$a^n + a^{n + 1} = a^{n + 2}; tag 1$



$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$



$a_pm = dfrac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2} = dfrac{1 pm sqrt 5}{2}; tag 3$



note that



$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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











3












$begingroup$

$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$






share|cite|improve this answer









$endgroup$









  • 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
















3












$begingroup$

$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$






share|cite|improve this answer









$endgroup$









  • 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














3












3








3





$begingroup$

$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$






share|cite|improve this answer









$endgroup$



$$1+a = a^2$$
$$text{By Quadratic formula, you get } a = frac {1 pm sqrt5}{2}$$
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 $a = frac {1 pm sqrt5}{2}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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










lollol is a new contributor. Be nice, and check out our Code of Conduct.










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