Sequences And Series Question 5
Question 5 - 25 January - Shift 2
For the two positive numbers $a, b$, if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then, $16 a+12 b$ is equal to
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Answer: 3
Solution:
Formula: Important Note (ii)
$a, b, \frac{1}{18} \to GP$
$\frac{a}{18}=b^{2} \ldots \ldots (i)$
$\frac{1}{a}, 10, \frac{1}{b} \to AP$
$\frac{1}{a}+\frac{1}{b}=20$
$\Rightarrow a+b=20 ab \ldots \ldots (ii)$
From eq. (i); we get,
$\Rightarrow 18 b^{2}+b=360 b^{3}$
$\Rightarrow 360 b^{2}-18 b-1=0 \quad{\because b \neq 0}$
$\Rightarrow b=\frac{18 \pm \sqrt{324+1440}}{720}$
$\Rightarrow b=\frac{18+\sqrt{1764}}{720} \quad{\because b>0}$
$\Rightarrow b=\frac{1}{12}$
$\Rightarrow a=18 \times \frac{1}{144}=\frac{1}{8}$
Now, $16 a+12 b=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3$
