For the two positive numbers a,b if a,b and 118are in a geometric progression while 1a,10 and 1b are in anarithmetic progression then 16a+12b is equal to…………….

a,b,118 are in GP⇒b2=a18⇒a=18b2 1a,10,1b are in AP⇒1a+1b=20⇒118b2+1b=20 ⇒360b2−18b−1=0⇒(12b−1)(30b+1)=0⇒b=112,b=−130(b>0)From 1, a=18⋅1144=18⇒16a+12b=2+1=3

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For the two positive numbers $a, b$ if $a, b a n d \frac{1}{18}$ are in a geometric progression while $\frac{1}{a}, 10 a n d \frac{1}{b}$ are in an
arithmetic progression then $16 a + 12 b$ is equal to…………….

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answer is 3.

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Detailed Solution

$a, b, \frac{1}{18} a r e i n G P \Rightarrow b^{2} = \frac{a}{18} \Rightarrow a = 18 b^{2} \frac{1}{a}, 10, \frac{1}{b} a r e i n AP \Rightarrow \frac{1}{a} + \frac{1}{b} = 20 \Rightarrow \frac{1}{18 b^{2}} + \frac{1}{b} = 20 \begin{array}{l} \Rightarrow 360 b^{2} - 18 b - 1 = 0 \Rightarrow (12 b - 1) (30 b + 1) = 0 \\ \Rightarrow b = \frac{1}{12}, b = \frac{- 1}{30} (b > 0) \end{array}$
From 1, $a = 18 \cdot \frac{1}{144} = \frac{1}{8} \Rightarrow 16 a + 12 b = 2 + 1 = 3$

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