Can someone help












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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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  • 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
















0












$begingroup$


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














0












0








0


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






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asked 3 hours ago









lollollollol

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181




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lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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










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