Question

If we have an equation \[4{a^2} + 9{b^2} + 16{c^2} = 2\left( {3ab + 6bc + 4ca} \right)\] , where \[a\] , \[b\]and \[c\] are non - zero numbers, then \[a\] , \[b\]and \[c\] are in
\[\left( 1 \right)\] \[A.P.\]
\[\left( 2 \right)\] \[G.P.\]
\[\left( 3 \right)\] \[H.P.\]
\[\left( 4 \right)\] None of these

Answer 1

Hint: We have to find the relation between \[a\] , \[b\]and \[c\] . We have to compute the relation between these three variables using the given expression . We should have the knowledge of the concept of arithmetic progression (A.P.) , geometric progression (G.P.) , harmonic progression (H.P.) . We should also have the concept of the mean of the progressions . Solving the expression we can compute the relation between \[a\] , \[b\]and \[c\] .

Complete step-by-step solution:
For the terms of a given series to be in G.P. the common ratio between the terms of the series should be the same for all the two consecutive terms of the series . The ratio of the second term to the first term of the given series should be the same as that of the ratio of the third term to the second term of the given series .
Given : \[4{a^2} + 9{b^2} + 16{c^2} = 2\left( {3ab + 6bc + 4ca} \right)\]
Multiplying both sides by \[2\] , we get
\[2 \times \left[ {4{a^2} + 9{b^2} + 16{c^2}} \right] = 2 \times 2\left( {3ab + 6bc + 4ca} \right)\]
Expanding the terms , we can write the expression as :
\[8{a^2} + 18{b^2} + 32{c^2} - \left( {12ab + 24bc + 16ca} \right) = 0\]
\[4{a^2} + 9{b^2} + 16{c^2} + 4{a^2} + 9{b^2} + 16{c^2} - \left( {12ab + 24bc + 16ca} \right) = 0\]
\[\left(4{a^2} -12ab+ 9{b^2}\right) + \left(9{b^2}-24bc + 16{c^2} \right) + \left(16{c^2} -16ca+ 4{a^2} \right) = 0\]
Using the term of square , using the formula
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
On simplifying , we get
\[{\left( {2a - 3b} \right)^2} + {\left( {3b - 4c} \right)^2} + {\left( {4c - 2a} \right)^2} = 0\]
As the sum of the square terms can’t be zero . As the square of any number is positive , the above equation is only possible when each square term is zero .
So ,
\[\left( {2a - 3b} \right) = 0\] or \[\left( {3b - 4c} \right) = 0\] or \[\left( {4c - 2a} \right) = 0\]
\[2a = 3b = 4c\]
Let ,us consider that the expression is as :
\[2a = 3b = 4c = x\]
\[a = \dfrac{x}{2}\] , \[b = \dfrac{x}{3}\] , \[c = \dfrac{x}{4}\]
Using the formula of harmonic mean for \[a\] , \[b\] , \[c\] we get
\[\dfrac{1}{c} + \dfrac{1}{a} = \dfrac{2}{b}\]
Putting the values in the above formula , we get the expression as :
\[\dfrac{2}{x} + \dfrac{4}{x} = \dfrac{6}{x}\]
Thus \[a\] , \[b\] and \[c\] are in a harmonic progression (H.P.) .
Hence , the correct option is \[\left( 3 \right)\] .

Note: We directly checked the condition for H.P. as while simplifying the given expression we got the values similar to the values which were satisfying the conditions of H.P. . Also, we don’t need to check for others as from the given options we get that only one of the options would be correct only . This is a point of smart work as it reduces the time taken to solve the question . also we can state that the given expressions are in A.P , G.P. or H.P. if it satisfies the conditions given below .
The formula of mean of the three progression is given as :
\[\left( 1 \right)\] \[A.P.\]
Arithmetic mean \[ = \left( {\dfrac{{a + b}}{2}} \right)\]
\[\left( 2 \right)\] \[G.P.\]
Geometric mean \[ = \sqrt {b \times c} \]
\[\left( 3 \right)\] \[H.P.\]
Harmonic mean \[ = \left( {\dfrac{{2ab}}{{a + b}}} \right)\]