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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 ab?
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in HP, where $\mathrm{p}^{-1}+\mathrm{q}^{-1}+\mathrm{r}^{-1}=\frac{5}{3}$
What is the value of ab?
Solution:
2976 Upvotes
Verified Answer
The correct answer is:
9
$\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(a+l)$
$\Rightarrow \mathrm{a}+\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{b}=\frac{5}{2}(a+b)$
$a+b+15=\frac{5}{2}(a+b)$
$\Rightarrow \mathrm{a}+\mathrm{b}=10$ ...(1)
$\& \frac{1}{a}+\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{b}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$
$\Rightarrow \quad \frac{1}{a}+\frac{1}{b}+\frac{5}{3}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$
$\Rightarrow \frac{3(a+b)}{a b}=\frac{10}{3}$ ...(2)
$\Rightarrow \quad \frac{3 \times 10}{a b}=\frac{10}{3}$
$\Rightarrow \quad a b=9$
$\Rightarrow \mathrm{a}+\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{b}=\frac{5}{2}(a+b)$
$a+b+15=\frac{5}{2}(a+b)$
$\Rightarrow \mathrm{a}+\mathrm{b}=10$ ...(1)
$\& \frac{1}{a}+\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{b}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$
$\Rightarrow \quad \frac{1}{a}+\frac{1}{b}+\frac{5}{3}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$
$\Rightarrow \frac{3(a+b)}{a b}=\frac{10}{3}$ ...(2)
$\Rightarrow \quad \frac{3 \times 10}{a b}=\frac{10}{3}$
$\Rightarrow \quad a b=9$
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