Sequences and Series 5 Question 17
8. Let $a$ and $b$ be positive real numbers. If $a, A _1, A _2, b$ are in arithmetic progression, $a, G _1, G _2, b$ are in geometric progression and $a, H _1, H _2, B$ are in harmonic progression, then show that
$$ \frac{G _1 G _2}{H _1 H _2}=\frac{A _1+A _2}{H _1+H _2}=\frac{(2 a+b)(a+2 b)}{9 a b} $$
(2002, 5M)
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Solution:
- Since, $a, A _1, A _2, b$ are in AP.
$$ \begin{array}{ll} \Rightarrow \quad & A _1+A _2=a+b \\ & a, G _1, G _2, b \text { are in GP } \quad \Rightarrow \quad G _1 G _2=a b \end{array} $$
and $\quad a, H _1, H _2, b$ are in HP.
$$ \begin{array}{ll} \Rightarrow & H _1=\frac{3 a b}{2 b+a}, H _2=\frac{3 a b}{b+2 a} \\ \therefore & \frac{1}{H _1}+\frac{1}{H _2}=\frac{1}{a}+\frac{1}{b} \\ \Rightarrow & \frac{H _1+H _2}{H _1 H _2}=\frac{A _1+A _2}{G _1 G _2}=\frac{1}{a}+\frac{1}{b} \end{array} $$
Now,
$$ \begin{aligned} \frac{G _1 G _2}{H _1 H _2} & =\frac{a b}{\frac{3 a b}{2 b+a} \frac{3 a b}{b+2 a}} \\ & =\frac{(2 a+b)(a+2 b)}{9 a b} \end{aligned} $$
From Eqs. (i) and (ii), we get
$$ \frac{G _1 G _2}{H _1 H _2}=\frac{A _1+A _2}{H _1+H _2}=\frac{(2 a+b)(a+2 b)}{9 a b} $$
