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Let $\mathrm{a}, \mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{b}$ be in $\mathrm{AP}$, where $\mathrm{x}+\mathrm{y}+\mathrm{z}=15$. Let $\mathrm{a}, \mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{b}$ be
in HP, where $\mathrm{p}^{-1}+\mathrm{q}^{-1}+\mathrm{r}^{-1}=\frac{5}{3}$
What is the value of $x y z$ ?
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in HP, where $\mathrm{p}^{-1}+\mathrm{q}^{-1}+\mathrm{r}^{-1}=\frac{5}{3}$
What is the value of $x y z$ ?
Solution:
2150 Upvotes
Verified Answer
The correct answer is:
105
On solving eq (1) \& (2), we get
(i) $a=1 \& b=9 \Rightarrow a+4 d=9 \Rightarrow d=2$
(ii) $a=9 \& b=1 \Rightarrow a+4 d=1 \Rightarrow d=-2$.
For $a=1 \& d=2$, $x=3, y=5 \& z=7$
For $a=9 \& d=-2$
$x=7, y=5 \& z=3$
$\Rightarrow \quad x y z=7 \times 5 \times 3=105$
(i) $a=1 \& b=9 \Rightarrow a+4 d=9 \Rightarrow d=2$
(ii) $a=9 \& b=1 \Rightarrow a+4 d=1 \Rightarrow d=-2$.
For $a=1 \& d=2$, $x=3, y=5 \& z=7$
For $a=9 \& d=-2$
$x=7, y=5 \& z=3$
$\Rightarrow \quad x y z=7 \times 5 \times 3=105$
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