Q.The AM of two numbers exceeds their GM by 10 and HM by 16. Find the numbers. OR If $P_1$ and $P_2$ are the lengths of the perpendiculars from the origin to the straight lines $x\sec\theta + y\,\text{cosec}\,\theta = 2a$ and $x\cos\theta - y\sin\theta = a\cos 2\theta$, then prove that $P_1^2 + P_2^2 = a^2$.

Setting AM $=M$, GM $=M-10$, HM $=M-16$, the identity $\text{GM}^2=\text{AM}\cdot\text{HM}$ gives $M=25$, and then solving $a+b=50,\ ab=225$ gives the numbers 45 and 5.

Let the two numbers be $a,b$. Let $M=\text{AM}=\dfrac{a+b}{2}$.

Given: $\text{GM} = M-10$ and $\text{HM}=M-16$.

For any two positive numbers, $\text{GM}^2 = \text{AM}\times\text{HM}$ (a standard identity: $\sqrt{ab}\cdot\sqrt{ab} = \frac{a+b}{2}\cdot\frac{2ab}{a+b}=ab$). So:

$(M-10)^2 = M(M-16)$

$M^2-20M+100 = M^2-16M$

$-20M+100=-16M \implies -4M=-100 \implies M=25$

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