Question

If ${a^2} + {b^2},ab + bc{\text{ and }}{b^2} + {c^2}$ are in G.P, prove that a , b , c are also in G.P.

Answer 1

Hint: We know when a , b , c are in G.P. then ${b^2} = ac$ it means when we prove ${b^2} = ac$ then we can clearly say a , b, c are in GP so to prove we have to use the data given in question.

Complete step-by-step answer:
We have given
${a^2} + {b^2},ab + bc{\text{ and }}{b^2} + {c^2}$ are in GP
And we know when three terms are in GP then square of second term will be equal to multiply of first term and third term.
So we can write
${\left( {ab + bc} \right)^2} = \left( {{a^2} + {b^2}} \right)\left( {{b^2} + {c^2}} \right)$
Now expanding we get,
${a^2}{b^2} + {b^2}{c^2} + 2a{b^2}c = {a^2}{b^2} + {a^2}{c^2} + {b^4} + {b^2}{c^2}$
On cancel out we get,
$2a{b^2}c = {a^2}{c^2} + {b^4}$
Now we can rewrite it as :
$
  {a^2}{c^2} + {b^4} - 2a{b^2}c = 0 \\
  {\left( {{b^2} - ac} \right)^2} = 0 \\
$
On taking square root on both side we get,
$
{\ {b^2} - ac} = 0 \\
$
So we get,
${b^2} = ac$ it means a , b, c are in GP where b is second term because we know when three terms are in GP then square of second term will be equal to multiply of first term and third term.
Hence a , b, c in GP proved.

Note: Whenever we get this type of question the key concept of solving is we have to start from given and use the result of GP that when three terms are in GP then square of second term will be equal to multiply of first term and third term. Using this we will be able to prove a, b, c in GP.