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Let $a,b,c$ be three consecutive terms of an arithmetic progression and a geometric one($a \neq 0$).Find the value of: $$\frac{a^3b+2b^2c^2-c^4}{a^2b^2+b^4-3c^4}$$

My attempts:Letting $a=b=c$ shows us that the answer is $-2$.Writing $a,b,c$ using the geometric progression $a,aq,aq^2$ we will get into an expression by $q$ which I don't know how to finish from here knowing that $a,b,c$ form an arithmetic progression.

J. W. Tanner
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Taha Akbari
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    If $a,b,c$ are an arithmetic sequence, then $b=\frac{a+c}{2}.$ If they are geometric, then $b^2=ac.$ Show that $\left(\frac{a+c}{2}\right)^2=ac$ only when $a=c.$ – Thomas Andrews Oct 29 '19 at 18:46
  • @ThomasAndrews Wow.I don't know how I missed that.Thank you for oppening my eyes. – Taha Akbari Oct 29 '19 at 18:49
  • On the other hand, if order is not important, then $-\frac{1}{2},\frac{1}{4},1$ is an arithmetic progression, and $\frac{1}{4},-\frac{1}{2},1$ are in arithmetic progression. – Thomas Andrews Oct 29 '19 at 19:10

2 Answers2

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We have given $$a,b=aq,c=aq^2$$ and $$a,b=a+d,c=a+2d$$ so we get $$aq=a+d$$ and $$aq^2=a+2d$$ eliminating $d$ we get $$aq^2=2aq-a$$ since $$a\neq 0$$ we get for $q$ the equation $$(q-1)^2=0$$ Can you finish?

David C. Ullrich
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Dr. Sonnhard Graubner
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If $a,b,c$ are in arithmetic progression, then $b-a=c-b$ or $2b=a+c.$

If $a,b,c$ are in geometric expression, then $\frac{b}{a}=\frac{c}{b},$ or $b^2=ac.$

This means $(a+c)^2=4ac.$ Show that if $(a+c)^2=4ac,$ then $a=c.$

Thomas Andrews
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