Question

If $a,b$ and $c$ are in A.P., and $p$ and $p'$ are respectively A.M. and G.M. between $a$ and $b$ while $q,q'$ are respectively the A.M. and G.M. between $b$ and $c$, then
A. ${{p}^{2}}+{{q}^{2}}=p{{'}^{2}}+q{{'}^{2}}$
B. $pq=p'q'$
C. ${{p}^{2}}-{{q}^{2}}=p{{'}^{2}}-q{{'}^{2}}$
D. None of these

Answer 1

Hint: In this question, we are to find the relation between the given arithmetic means and geometric means of the terms $a,b$ and $c$ which are in A.P. For this, by applying appropriate mean formulae, we get the required expression.

Formula used: If $a,b$ and $c$ are in arithmetic progression, then their arithmetic mean is
$A.M=2b=a+c$
Here arithmetic mean is nothing but the average between the first and last term i.e., nothing but the middle term.
If $a,b$, and $c$ are in geometric progression, then their geometric mean is
$b=\sqrt{ac}$
Here the geometric mean is nothing but the square of the middle term. I.e., obtained by multiplying the first and last terms.

Complete step by step solution: Given that,
$a,b$ and $c$ are in arithmetic progression, then their arithmetic mean is
$2b=a+c\text{ }...(1)$
It is given that $p$ and $q$ are the arithmetic means between $a,b$ and $b,c$.
So, from the arithmetic mean definition, we can write the series as
$a,p,b,q,c$
Thus, we can write
$p=\dfrac{a+b}{2}$ and $q=\dfrac{b+c}{2}$
Now, squaring these two means on both sides, we get
${{p}^{2}}=\dfrac{{{(a+b)}^{2}}}{4}\text{ }...(2)$
And
${{q}^{2}}=\dfrac{{{(b+c)}^{2}}}{4}\text{ }...(3)$
It is also given that, $p'$ and $q'$ are the geometric means between $a,b$ and $b,c$.
So, from the geometric mean definition, we can write the series as
$a,p',b,q',c$
Thus, we can write
$\begin{align}
  & p'=\sqrt{ab} \\
 & \Rightarrow p{{'}^{2}}=ab\text{ }...(4) \\
\end{align}$
$\begin{align}
  & q{{'}^{2}}=\sqrt{bc} \\
 & \Rightarrow q{{'}^{2}}=bc\text{ }...(5) \\
\end{align}$
Then, from (2) and (3),
$\begin{align}
  & {{p}^{2}}-{{q}^{2}}=\dfrac{{{(a+b)}^{2}}}{4}-\dfrac{{{(b+c)}^{2}}}{4} \\
 & \text{ }=\dfrac{{{(a+b)}^{2}}-{{(b+c)}^{2}}}{4} \\
 & \text{ }=\dfrac{{{a}^{2}}+2ab+{{b}^{2}}-{{b}^{2}}-2bc-{{c}^{2}}}{4} \\
 & \text{ }=\dfrac{({{a}^{2}}-{{c}^{2}})+2ab-2bc}{4} \\
 & \text{ }=\dfrac{(a-c)(a+c)+2ab-2bc}{4} \\
\end{align}$
On substituting (1), we get
$\begin{align}
  & {{p}^{2}}-{{q}^{2}}=\dfrac{2b(a-c)+2ab-2bc}{4} \\
 & \text{ }=\dfrac{2ab-2bc+2ab-2bc}{4} \\
 & \text{ }=\dfrac{4(ab-bc)}{4} \\
 & \text{ }=(ab-bc) \\
\end{align}$
On substituting (4) and (5) in the above, we get
\[\begin{align}
  & {{p}^{2}}-{{q}^{2}}=ab-bc=p{{'}^{2}}-q{{'}^{2}} \\
 & \therefore {{p}^{2}}-{{q}^{2}}=p{{'}^{2}}-q{{'}^{2}} \\
\end{align}\]

Thus, Option (C) is correct.

Note: Here we need to remember that, the means represent the middle terms of the series. So, by using this concept, the above-given question is solved.