Question

For any three positive real numbers a, b and c, $ 9(25{a^2} + {b^2}) + 25({c^2} - 3ac) = 15b(3a + c) $ , then.
A.b, c and a are in G.P.
B.b, c and a are in A,P.
C.a, b and c are in A.P.
D.a, b and c are in G.P.

Answer 1

Hint: First, simplify the equation and find out the relation between a, b and c. Then apply the conditions for a given set of numbers to be in A.P. and G.P. and check which option verifies the conditions to get the correct answer. Condition for a given set of numbers to be in G.P. is that it should verify the geometric mean and that for A.P. is that it should verify the arithmetic mean.

Complete step-by-step answer:
Let us start by simplifying the given equation,
 $
  9(25{a^2} + {b^2}) + 25({c^2} - 3ac) = 15b(3a + c) \\
  225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc \\
  225{a^2} + 9{b^2} + 25{c^2} - 75ac - 45ab - 15bc = 0 \;
  $
Multiply both sides by 2
 $
  2(225{a^2} + 9{b^2} + 25{c^2} - 75ac - 45ab - 15bc) = 2(0) \\
  450{a^2} + 18{b^2} + 50{c^2} - 150ac - 90ab - 30bc = 0 \;
  $
Now rearranging the above equation, we get –
 $ 225{a^2} + 9{b^2} - 90ab + 9{b^2} + 25{c^2} - 30bc + 225{a^2} + 25{c^2} - 150ac = 0 $
Expand the numerical constants in the above equation,
 $ {(15a)^2} + {(3b)^2} - 2 \times 15a \times 3b + {(3b)^2} + {(5c)^2} - 2 \times 3b \times 5c + {(15a)^2} + {(5c)^2} - 2 \times 15a \times 5c = 0 $
Now we know that, $ {(a - b)^2} = {a^2} + {b^2} - 2ab $ , use this identity in the above equation.
 $ {(15a - 3b)^2} + {(3b - 5c)^2} + {(15a - 5c)^2} = 0 $
From the above equation, we get –
 $
  15a = 3b \\
   \Rightarrow a = \dfrac{b}{5} \\
  3b = 5c \\
   \Rightarrow c = \dfrac{{3b}}{5} \\
  15a = 5c \\
   \Rightarrow c = 3a \;
  $
If b, c and a are in G.P. then $ {c^2} = ba $
Here $ {c^2} = \dfrac{{5c}}{3} \times \dfrac{c}{3} = \dfrac{{5{c^2}}}{9} $
The Left-Hand side is not equal to the Right-Hand side, thus b, c and a are not in G.P.
If b, c and a are in A.P. then $ c = \dfrac{{a + b}}{2} $
Here $ c = \dfrac{{\dfrac{c}{3} + \dfrac{{5c}}{3}}}{2} = \dfrac{{6c}}{6} = c $
The Left-Hand side is equal to the Right-Hand side.
Thus b, c and a are in A.P.
We don’t need to check the rest of the options as we have got the correct answer.
So, the correct answer is “Option B”.

Note: A geometric progression is a sequence of numbers in which each term is found by the multiplication of its previous term with a non-one number, this non-one number is called the common ratio.
An arithmetic progression is a progression or sequence of numbers such that the difference between any two consecutive numbers is constant.