Suppose a,b,c are in A.P.

Question

Suppose $a,b,c$ are in A.P. and ${a}^{2},2{b}^{2},{c}^{2}$ are in G.P. If $a<b<c$ and $a+b+c=1$, then $9({a}^{2}+{b}^{2}+{c}^{2})$ is equal to $____$ .

Answer: 9

Accepted Answer

$\begin{gathered} a=b-d,c=b+d,\Rightarrow b=\frac{1}{3}\Rightarrow 4{b}^{4}={a}^{2}{c}^{2} \ 4{b}^{4}=[(b-d)(b+d){]}^{2} \ \frac{4}{81}={(\frac{1}{9}-{d}^{2})}^{2}\Rightarrow \frac{4}{81}=\frac{1}{81}-\frac{2{d}^{2}}{9}+{d}^{4}\Rightarrow {d}^{4}-\frac{2{d}^{2}}{9}-\frac{1}{27}=0\Rightarrow 27{d}^{4}-6{d}^{2}-1=0 \ {d}^{2}=1/3\Rightarrow d=+\frac{1}{\sqrt{3}}(a>b>c) \ 9({a}^{2}+{b}^{2}+{c}^{2})=9[{(\frac{1}{3}-\frac{1}{\sqrt{3}})}^{2}+{(\frac{1}{3})}^{2}+{(\frac{1}{3}+\frac{1}{\sqrt{3}})}^{2}]=9[\frac{1}{3}+\frac{2}{3}]=3+6=9 \end{gathered}$

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