The first three terms of a geometric progress | vedclass

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(B) Let the three terms be $\frac{b}{r}, b, br$ where $r$ is the common ratio.Given that the harmonic mean of $a$ and $b$ is $12$:$\frac{2ab}{a+b} = 1

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let the three terms be $\frac{b}{r}, b, br$ where $r$ is the common ratio.Given that the harmonic mean of $a$ and $b$ is $12$:$\frac{2ab}{a+b} = 12 \implies \frac{2(\frac{b}{r})b}{\frac{b}{r}+b} = 12 \implies \frac{2b}{1+r} = 12 \implies b = 6(1+r) \dots (1)$Given that the harmonic mean of $b$ and $c$ is $36$:$\frac{2bc}{b+c} = 36 \implies \frac{2b(br)}{b+br} = 36 \implies \frac{2br}{1+r} = 36 \implies br = 18(1+r) \dots (2)$Dividing $(2)$ by $(1)$:$\frac{br}{b} = \frac{18(1+r)}{6(1+r)} \implies r = 3$Substituting $r=3$ into $(1)$:$b = 6(1+3) = 24$Since $a = \frac{b}{r} = \frac{24}{3} = 8$.

The first three terms of a geometric progression are $a, b, c$. If the harmonic mean of $a$ and $b$ is $12$ and the harmonic mean of $b$ and $c$ is $36$,then $a = \dots$

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