JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 18)
Let $${1 \over {16}}$$, a and b be in G.P. and $${1 \over a}$$, $${1 \over b}$$, 6 be in A.P., where a, b > 0. Then 72(a + b) is equal to ___________.
Answer
14
Explanation
$${a^2} = {b \over {16}}$$ and $${2 \over b} = {1 \over a} + 6$$
Solving, we get $$a = {1 \over {12}}$$ or $$a = - {1 \over 4}$$ [rejected]
if $$a = {1 \over {12}} \Rightarrow b = {1 \over 9}$$
$$ \therefore $$ $$72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$$
Solving, we get $$a = {1 \over {12}}$$ or $$a = - {1 \over 4}$$ [rejected]
if $$a = {1 \over {12}} \Rightarrow b = {1 \over 9}$$
$$ \therefore $$ $$72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$$
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