For the two positive numbers a , b , if a , b and 1 18 are in a geometric progression, while 1 a , 10 and 1 b are in an arithmetic progression, then 16 a + 12 b is equal to ________ . [2023]

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Solution

Q.
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 ________ . [2023]

Ans.
(3)

$a, b, \frac{1}{18}$ are in G.P. $\Rightarrow b^{2} = \frac{a}{18}$

$\Rightarrow a = 18 b^{2}$ $...(i)$

$\frac{1}{a}, 10, \frac{1}{b} are in A.P. \Rightarrow 20 = \frac{1}{a} + \frac{1}{b}$

$\Rightarrow a + b = 20 a b \Rightarrow 18 b^{2} + b = 20 \cdot 18 b^{2} b [using (i)]$

$\Rightarrow 18 b + 1 = 360 b^{2} \Rightarrow 360 b^{2} - 18 b - 1 = 0$

$\Rightarrow 360 b^{2} - 30 b + 12 b - 1 = 0 \Rightarrow (30 b + 1) (12 b - 1) = 0$

$which gives b = \frac{1}{12} (Taking only positive value)$

$So, a = \frac{1}{8} .$

$Hence, 16 a + 12 b = 16 \times \frac{1}{8} + 12 \times \frac{1}{12} = 2 + 1 = 3 .$

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