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Let $a,b$ and $c$ be consecutive numbers in a geometric sequence, where $a+b\neq 0$ and $b+c\neq 0$. Show that $$ \frac{2ab}{a+b}, b, \frac{2bc}{b+c} $$ are consecutive terms in an arithmetic sequence.

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2 Answers 2

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Since $a$, $b$, $c$ are consecutive numbers in a geometric sequence, there exist a real number $r$ such that $b=ar$ and $c=ar^2$. Since $a+b\neq0$ and $b+c\neq0$, it follows $a\neq0$, $r\neq-1$, and $r\neq 0$; then observe \begin{align} b-\dfrac{2ab}{a+b}&=\frac{ab+b^2-2ab}{a+b}=\frac{b^2-ab}{a+b}=\frac{b(b-a)}{a+b}=\frac{ar(ar-a)}{a+ar}=\frac{ar(r-1)}{1+r} \end{align} On the other hand, $$\dfrac{2bc}{b+c}-b=\frac{2bc-b^2-bc}{b+c}=\frac{bc-b^2}{b+c}=\frac{b(c-b)}{b+c}=\frac{ar(ar^2-ar)}{ar+ar^2}=\frac{ar(r-1)}{1+r}$$ From this the result follows.

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Another method, which is no easier than Ángel Mario Gallegos's answer but in another case might well be more elegant:

Make the same starting observations as in his:

Since $a$, $b$, $c$ are consecutive numbers in a geometric sequence, there exist a real number $r$ such that $b=ar$ and $c=ar^2$. Since $a+b\neq0$ and $b+c\neq0$, it follows $a\neq0$, $r\neq-1$, and $r\neq > 0$.

Then take the mean of the first and third terms:

\begin{align} \frac{1}{2}\left(\frac{2ab}{a+b}+\frac{2bc}{b+c}\right)&=\frac{ab}{a+b}+\frac{bc}{b+c}\\ &=\frac{a^2r}{a+ar}+\frac{a^2r^3}{ar+ar^2}\\ &=\frac{ar}{1+r}+\frac{ar^2}{1+r}\\ &=\frac{ar(1+r)}{1+r}\\ &=ar\\ &=b \end{align}

Since the second term is midway between the first and third, the three must form consequtive terms of an arithmetic sequence.

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