Let 1 16 , a and b be in G.P. and 1 a , 1 b , 6 be in A.P., where a , b > 0 . Then 72 ( a + b ) is equal to _______ .

a 2 = b 16 ⇒ 1 b = 1 16 a 2 2 b = 1 a + 6 1 8 a 2 = 1 a + 6 1 a 2 - 8 a - 48 = 0 1 a = 12 , - 4 ⇒ a = 1 12 , - 1 4 a = 1 12 , a > 0 b = 16 a 2 = 1 9 ⇒ 72 ( a + b ) = 6 + 8 = 14

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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesMiscellaneous/MixedEasy2 minPYQ_2021

MathematicsEasystatement

Let $\frac{1}{16}, a$ and $b$ be in G.P. and $\frac{1}{a}, \frac{1}{b}, 6$ be in A.P., where $a, b > 0$ . Then $72 (a + b)$ is equal to _______ .

Answer:

Solution:

$a^{2} = \frac{b}{16} \Rightarrow \frac{1}{b} = \frac{1}{16 a^{2}}$

$\frac{2}{b} = \frac{1}{a} + 6$

$\frac{1}{8 a^{2}} = \frac{1}{a} + 6$

$\frac{1}{a^{2}} - \frac{8}{a} - 48 = 0$

$\frac{1}{a} = 12, - 4 \Rightarrow a = \frac{1}{12}, - \frac{1}{4}$

$a = \frac{1}{12}, a > 0$

$b = 16 a^{2} = \frac{1}{9}$

$\Rightarrow 72 (a + b) = 6 + 8 = 14$

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2021

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