Question

The point $P(a,b)$ is such that $b-25a=4$ and the arithmetic mean of $a$ an $b$ is $28$. $Q(x,y)$ is the point such that $x$ and $y$ are two geometric means between $a$ and $b$ if $O$ is the origin then $O{P}^2+O{Q}^2$ is equal to

Answer 1

Given that $b-25a=4$ ...(1)
Also, $a+b=28\times 2=56$ ...(2)
Subtracting equation 1 from 2, we get
$a+b-b+25a=56-4=52$
i.e. $a=2$, and substituting $a=2$ in equation 1, we get b to be $54$.
Now, $x$ and $y$ are 2 geometric means between $2$ and $54$, so $x=2r,y=2r^2$ and $54=2r^3$
Thus, $r={27}^{\frac{1}{3}}=3$
Hence, $x=6$ and $y=18$
$\therefore P=(2,54),Q=(6,18),O=(0,0)$
$O{P}^2+O{Q}^2=2^2+{54}^2+6^2+{18}^2=4+2916+36+324=3280$