Let a, b, c be in A.P.

Question

Let $a, b, c$ be in A.P. such that $a^2, 2b^2, c^2$ are in G.P., with $a < b < c$ and $a + b + c = 1$. Then $9(a^2 + b^2 + c^2)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 9

Accepted Answer

Since $a, b, c$ are in A.P., write $a = b-d$, $c = b+d$. Also $a+b+c=1 \Rightarrow 3b=1 \Rightarrow b = \tfrac13$.

Since $a^2, 2b^2, c^2$ are in G.P., $4b^4 = a^2c^2 = [(b-d)(b+d)]^2 = (b^2-d^2)^2$.

$$4b^4 = \left(\frac19 - d^2\right)^2 \implies \frac{4}{81} = \frac{1}{81} - \frac{2d^2}{9} + d^4 \implies 27d^4 - 6d^2 - 1 = 0.$$

Solving this quadratic in $d^2$: $d^2 = \tfrac13$, so $d = \tfrac{1}{\sqrt3}$ (taking the sign consistent with $a<b<c$).

$$9(a^2+b^2+c^2) = 9\left[\left(\frac13 - \frac{1}{\sqrt3}\right)^2 + \left(\frac13\right)^2 + \left(\frac13+\frac{1}{\sqrt3}\right)^2\right] = 9\left[\frac13 + \frac23\right] = 9.$$

Solutions · 1