If a, b, c are in G.P then
A. \[{{\text{a}}^{\text{2}}}{\text{,b}}{}^{\text{2}}{\text{,}}{{\text{c}}^{\text{2}}}\]are in G.P
B. \[{{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),b}}{}^{\text{2}}{\text{(a + c)}}\]are in G.P
C. \[\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{a + c}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}\]are in G.P
D. None of these

Answer Verified Verified
Hint: Using the given concept of G.P, \[{{\text{b}}^{\text{2}}}{\text{ = ac}}\],as known that the product of first and last term is equal to square of the middle term. Hence, using this and applying trial and error method to calculate the above question.

Complete step by step answer:

As given that a, b, c are in G.P
As we know that \[{{\text{b}}^{\text{2}}}{\text{ = ac}}\]
Hence on squaring both the side of the above equation we can get
  {{\text{(}}{{\text{b}}^{\text{2}}}{\text{)}}^{\text{2}}}{\text{ = (ac}}{{\text{)}}^{\text{2}}} \\
  {{\text{(}}{{\text{b}}^{\text{2}}}{\text{)}}^{\text{2}}}{\text{ = (}}{{\text{a}}^{\text{2}}}{\text{)(}}{{\text{c}}^{\text{2}}}{\text{)}} \\
Hence, \[{{\text{a}}^{\text{2}}}{\text{,b}}{}^{\text{2}}{\text{,}}{{\text{c}}^{\text{2}}}\]are in G.P
And so option(a) is our correct answer.

Note: A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the previous term is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
Sequences are lists of numbers placed in a definite order according to given rules. The series corresponding to a sequence is the sum of the numbers in that sequence. Series can be arithmetic, meaning there is a fixed difference between the numbers of the series, or geometric, meaning there is a fixed factor.