Prove a geometric sequence a, b, c from the arithmetic progression $1/(b-a)$, $1/2b$, $1/(b-c)$
$begingroup$
The given task is:
The following forms an arithmetic sequence: $$frac{1}{b-a}, frac{1}{2b}, frac{1}{b-c}.$$
Show, that $a, b, c$ forms an geometric sequence.
It's easily enough to understand that $$ frac{1}{2b}-frac{1}{b-a}=frac{1}{b-c}-frac{1}{2b} iff frac{a+b}{a-b}=frac{b+c}{b-c}$$ and that I need to prove that $$frac{b}{a}=frac{c}{b},$$ but between the two I just make it more and more complicated.
arithmetic-progressions geometric-progressions
asked Dec 3 '18 at 14:52
Anna LindebergAnna Lindeberg
204
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add a comment |
$begingroup$
The given task is:
The following forms an arithmetic sequence: $$frac{1}{b-a}, frac{1}{2b}, frac{1}{b-c}.$$
Show, that $a, b, c$ forms an geometric sequence.
It's easily enough to understand that $$ frac{1}{2b}-frac{1}{b-a}=frac{1}{b-c}-frac{1}{2b} iff frac{a+b}{a-b}=frac{b+c}{b-c}$$ and that I need to prove that $$frac{b}{a}=frac{c}{b},$$ but between the two I just make it more and more complicated.
arithmetic-progressions geometric-progressions
asked Dec 3 '18 at 14:52
Anna LindebergAnna Lindeberg
204
$endgroup$
$begingroup$
brilliant.org/wiki/componendo-and-dividendo
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 12:34
add a comment |
$begingroup$
The given task is:
The following forms an arithmetic sequence: $$frac{1}{b-a}, frac{1}{2b}, frac{1}{b-c}.$$
Show, that $a, b, c$ forms an geometric sequence.
It's easily enough to understand that $$ frac{1}{2b}-frac{1}{b-a}=frac{1}{b-c}-frac{1}{2b} iff frac{a+b}{a-b}=frac{b+c}{b-c}$$ and that I need to prove that $$frac{b}{a}=frac{c}{b},$$ but between the two I just make it more and more complicated.
arithmetic-progressions geometric-progressions
asked Dec 3 '18 at 14:52
Anna LindebergAnna Lindeberg
204
$endgroup$
The given task is:
The following forms an arithmetic sequence: $$frac{1}{b-a}, frac{1}{2b}, frac{1}{b-c}.$$
Show, that $a, b, c$ forms an geometric sequence.
It's easily enough to understand that $$ frac{1}{2b}-frac{1}{b-a}=frac{1}{b-c}-frac{1}{2b} iff frac{a+b}{a-b}=frac{b+c}{b-c}$$ and that I need to prove that $$frac{b}{a}=frac{c}{b},$$ but between the two I just make it more and more complicated.
arithmetic-progressions geometric-progressions
arithmetic-progressions geometric-progressions
asked Dec 3 '18 at 14:52
Anna LindebergAnna Lindeberg
204
asked Dec 3 '18 at 14:52
Anna LindebergAnna Lindeberg
204
asked Dec 3 '18 at 14:52
Anna LindebergAnna Lindeberg
204
asked Dec 3 '18 at 14:52
Anna LindebergAnna Lindeberg
204
asked Dec 3 '18 at 14:52
Anna LindebergAnna Lindeberg
204
204
$begingroup$
brilliant.org/wiki/componendo-and-dividendo
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 12:34
add a comment |
$begingroup$
brilliant.org/wiki/componendo-and-dividendo
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 12:34
$begingroup$
brilliant.org/wiki/componendo-and-dividendo
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 12:34
$begingroup$
brilliant.org/wiki/componendo-and-dividendo
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 12:34
add a comment |
2 Answers
2
active
oldest
votes
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Taking it from where you left off, use cross-products and simplify
$$(a+b)(b-c) = (a-b)(b+c) iff color{blue}{ab}-ac+b^2color{green}{-bc} = color{blue}{ab}+ac-b^2color{green}{-bc}$$
$$2b^2 = 2ac iff b^2 = ac iff frac{b}{a} = frac{c}{b}$$
answered Dec 3 '18 at 15:02
KM101KM101
6,0251525
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add a comment |
$begingroup$
We have
$$
frac1b = frac{1}{b-a} + frac1{b-c}\
(b-a)(b-c) = (b-c + b-a)b\
b^2 - ab - bc + ac = 2b^2-bc - ab\
ac = b^2
$$
and we are done.
answered Dec 3 '18 at 15:02
ArthurArthur
116k7116198
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Taking it from where you left off, use cross-products and simplify
$$(a+b)(b-c) = (a-b)(b+c) iff color{blue}{ab}-ac+b^2color{green}{-bc} = color{blue}{ab}+ac-b^2color{green}{-bc}$$
$$2b^2 = 2ac iff b^2 = ac iff frac{b}{a} = frac{c}{b}$$
answered Dec 3 '18 at 15:02
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
Taking it from where you left off, use cross-products and simplify
$$(a+b)(b-c) = (a-b)(b+c) iff color{blue}{ab}-ac+b^2color{green}{-bc} = color{blue}{ab}+ac-b^2color{green}{-bc}$$
$$2b^2 = 2ac iff b^2 = ac iff frac{b}{a} = frac{c}{b}$$
answered Dec 3 '18 at 15:02
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
Taking it from where you left off, use cross-products and simplify
$$(a+b)(b-c) = (a-b)(b+c) iff color{blue}{ab}-ac+b^2color{green}{-bc} = color{blue}{ab}+ac-b^2color{green}{-bc}$$
$$2b^2 = 2ac iff b^2 = ac iff frac{b}{a} = frac{c}{b}$$
answered Dec 3 '18 at 15:02
KM101KM101
6,0251525
$endgroup$
Taking it from where you left off, use cross-products and simplify
$$(a+b)(b-c) = (a-b)(b+c) iff color{blue}{ab}-ac+b^2color{green}{-bc} = color{blue}{ab}+ac-b^2color{green}{-bc}$$
$$2b^2 = 2ac iff b^2 = ac iff frac{b}{a} = frac{c}{b}$$
answered Dec 3 '18 at 15:02
KM101KM101
6,0251525
answered Dec 3 '18 at 15:02
KM101KM101
6,0251525
answered Dec 3 '18 at 15:02
KM101KM101
6,0251525
answered Dec 3 '18 at 15:02
KM101KM101
6,0251525
6,0251525
add a comment |
add a comment |
$begingroup$
We have
$$
frac1b = frac{1}{b-a} + frac1{b-c}\
(b-a)(b-c) = (b-c + b-a)b\
b^2 - ab - bc + ac = 2b^2-bc - ab\
ac = b^2
$$
and we are done.
answered Dec 3 '18 at 15:02
ArthurArthur
116k7116198
$endgroup$
add a comment |
$begingroup$
We have
$$
frac1b = frac{1}{b-a} + frac1{b-c}\
(b-a)(b-c) = (b-c + b-a)b\
b^2 - ab - bc + ac = 2b^2-bc - ab\
ac = b^2
$$
and we are done.
answered Dec 3 '18 at 15:02
ArthurArthur
116k7116198
$endgroup$
add a comment |
$begingroup$
We have
$$
frac1b = frac{1}{b-a} + frac1{b-c}\
(b-a)(b-c) = (b-c + b-a)b\
b^2 - ab - bc + ac = 2b^2-bc - ab\
ac = b^2
$$
and we are done.
answered Dec 3 '18 at 15:02
ArthurArthur
116k7116198
$endgroup$
We have
$$
frac1b = frac{1}{b-a} + frac1{b-c}\
(b-a)(b-c) = (b-c + b-a)b\
b^2 - ab - bc + ac = 2b^2-bc - ab\
ac = b^2
$$
and we are done.
answered Dec 3 '18 at 15:02
ArthurArthur
116k7116198
answered Dec 3 '18 at 15:02
ArthurArthur
116k7116198
answered Dec 3 '18 at 15:02
ArthurArthur
116k7116198
answered Dec 3 '18 at 15:02
ArthurArthur
116k7116198
116k7116198
add a comment |
add a comment |
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$begingroup$
brilliant.org/wiki/componendo-and-dividendo
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 12:34