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, $16 a+12 b=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3$