"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 ___________."

Question

"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 ___________."

Accepted Answer

"${a^2} = {b \over {16}}$ and ${2 \over b} = {1 \over a} + 6$

Solving, we get $a = {1 \over {12}}$ or $a = - {1 \over 4}$ [rejected]

if $a = {1 \over {12}} \Rightarrow b = {1 \over 9}$

$\therefore$ $72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$"

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 ___________.

Solution

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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 ___________.

Solution

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