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If a, b, c is in AP a, mb, c is in GP then a, ${{m}^{2}}b$, c is in?

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Last updated date: 23rd May 2024
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Answer
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Hint: In the given question we are given some terms of arithmetic progression and some terms of geometric progression and some more terms are also given which we need to identify what kind of series it belongs to or is any kind of progression for which we need to clearly use the definition of each terms.

Complete step by step answer:
According to the question, we are given that a, b, c is in AP and we know that in any arithmetic progression the common difference between the consecutive terms is the same. Therefore, we can write as
 $\begin{align}
  & b-a=c-b \\
 & \Rightarrow 2b=c-a \\
 & \Rightarrow b=\dfrac{c-a}{2} \\
\end{align}$
Now, similarly we know that in GP the common ratio between the consecutive terms is same therefore, we get,
 $\begin{align}
  & \dfrac{mb}{a}=\dfrac{c}{mb} \\
 & \Rightarrow {{m}^{2}}{{b}^{2}}=ac \\
\end{align}$
Now, making some substitutions in this we get,
$\begin{align}
  & {{m}^{2}}bb=ac \\
 & \Rightarrow {{m}^{2}}b\left( \dfrac{c-a}{2} \right)=ac \\
 & \Rightarrow {{m}^{2}}b=\dfrac{2ac}{c-a} \\
\end{align}$
And this clearly shows the given three terms are in HP as they satisfy the relation of harmonic series.
Therefore, a, ${{m}^{2}}b$, c is in HP.

Note: In such a type of question, we need to be careful with the definition of each type of progression or series involved. Also, we need to be aware of all the main terms involved and proceed directly as asked by the definition and always try to write the simplified answer and work step wise step in order to reduce chances of mistakes.