For the two positive numbers is and are in a geometric progression, while and are in an arithmetic progression, then is equal to

3 Explanation: Solving: GP: AP: (since ) Answer :- 3

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

For the two positive numbers $a, b$ is $a, b$ 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

Choose the correct answer:

A.
3
(Correct Answer)
B.
2
C.
1
D.
0

Correct Answer:

Explanation

Solving:

GP: $b^2 = \frac{a}{18} \implies a = 18b^2$

AP: $2(10) = \frac{1}{a} + \frac{1}{b} \implies 20 = \frac{1}{18b^2} + \frac{1}{b}$

$360b^2 = 1 + 18b \implies 360b^2 - 18b - 1 = 0$

$(30b + 1)(12b - 1) = 0 \implies b = \frac{1}{12}$ (since $b > 0$ )

$a = 18(\frac{1}{12})^2 = \frac{18}{144} = \frac{1}{8}$

$16a + 12b = 16(\frac{1}{8}) + 12(\frac{1}{12}) = 2 + 1 = 3$

Explanation

Solving:

GP: $b^2 = \frac{a}{18} \implies a = 18b^2$

AP: $2(10) = \frac{1}{a} + \frac{1}{b} \implies 20 = \frac{1}{18b^2} + \frac{1}{b}$

$360b^2 = 1 + 18b \implies 360b^2 - 18b - 1 = 0$

$(30b + 1)(12b - 1) = 0 \implies b = \frac{1}{12}$ (since $b > 0$ )

$a = 18(\frac{1}{12})^2 = \frac{18}{144} = \frac{1}{8}$

$16a + 12b = 16(\frac{1}{8}) + 12(\frac{1}{12}) = 2 + 1 = 3$

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