Question

If a, b, c are in AP, and p, $p^{{}^{\prime }}$ are in AM and GM respectively between a and b, while q, $q^{{}^{\prime }}$ are the AM and GM respectively between b and c, then

• A. $p^2+q^2=p^{{}^{{}^{\prime }2}}+q^{{}^{{}^{\prime }2}}$
• B. $pq=p^{{}^{\prime }}{q}^{{}^{\prime }}$
• C. $p^2-q^2=p^{{}^{{}^{\prime }2}}-q^{{}^{{}^{\prime }2}}$
• D. None of these

Answer 1

The correct option is C $p^2-q^2=p^{{}^{{}^{\prime }2}}-q^{{}^{{}^{\prime }2}}$

Given that, $a,b,c$ are in A.P
Therefore, $2b=a+c$ ...(1)
Since. $p$,$p^{‘}$ are A.M and G.M between $a$ & $b$ respectively
Therefore, $p=\frac{a+b}{2}$ and $p^{‘}=\sqrt{ab}$
Since. $q$,$q^{‘}$ are A.M and G.M between $b$ & $c$ respectively
Therefore, $q=\frac{b+c}{2}$ and $q^{‘}=\sqrt{bc}$
Consider, $(p+q)(p-q)-(p^{‘}+q^{‘})(p^{‘}-q^{‘})=\frac{(a+b+b+c)(a+b-b-c)}{4}-(\sqrt{ab}+\sqrt{bc})(\sqrt{ab}-\sqrt{bc})$
$\Rightarrow p^2-q^2-p^{{}^{\prime 2}}+q^{{}^{\prime 2}}=\frac{(a+c+2b)(a-c)}{4}-(ab-bc)$
$\Rightarrow p^2-q^2-p^{{}^{\prime 2}}+q^{{}^{\prime 2}}=\frac{(2b+2b)(a-c)}{4}-b(a-c)=0$ [ from (1) ]
Therefore, $p^2-q^2=p^{{}^{\prime 2}}-q^{{}^{\prime 2}}$