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$a, b,$ and $c$ numbers are making geometric progression, whereas $$(a + 2b), (2a + b + c), (a + 3b + c)$$ numbers are making arithmetic progression.

How can I find the $n$th term of the above G.P

NicolasDev153
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    Can you edit into your question what you have tried so far? – Henry Oct 12 '21 at 15:21
  • perhaps you would like to show us what you tried? – David Quinn Oct 12 '21 at 15:22
  • Incidentally, you seem to have two equations and three unknowns, and it looks as if you could rescale $a,b,c$ by a constant multiplicative factor, so I would expect the answer to be a function of $a$ and $n$ – Henry Oct 12 '21 at 15:24
  • $+1$ welcome to MSE though it's your first question on MSE it is expected to show the efforts you did to solve the problem. –  Oct 12 '21 at 16:25

1 Answers1

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$$a+2b, 2a+b+c, a+3b+c = u-d, u, u+d$$ $$\color{red}{(u-d)} + \color{blue}{u} + \color{green}{(u+d)} = 3\color{blue}{u} = \color{red}{(a+2b)}+\color{blue}{(2a+b+c)}+\color{green}{(a+3b+c)}$$ sum of three terms of AP : $3u = 4a+6b+2c = 3(2a+b+c) \implies \color{red}{3b = 2a+c}$

$a, b,c $ in G.P $$\implies \frac{c}{b} = \frac b{a} = r$$ $$\color{red}{r^2 = \frac c{a}}$$ $$\color{red}{b^2 =ac}$$ $$\left(\frac{2a+c}{3}\right)^2 = ac$$ $$\left(\frac{2+(\frac{c}{a})}{3}\right)^2 = \frac{c}{a} = r^2 = p$$ $$\left(\frac{2+p}3\right)^2 = p$$ $$\implies p = 1, 4$$ $$r = ±1, ±2$$