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Previous Year Question (PYQs)

Suppose $a,b,c$ are in A.P. and $a^2,2b^2,c^2$ are in G.P. If $a



Solution

Let $a=\frac{1}{3}-d,b=\frac{1}{3},c=\frac{1}{3}+d$. G.P. condition: $2b^2$ is geometric mean ⇒ $(2b^2)^2=a^2c^2$. Solving gives $d=\frac{1}{3}$. Then sum of squares gives $\frac{2}{3}$, so $9\times\frac{2}{3}=6$.

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