Suppose a, b, c are in A.P. and a^{2}, 2b^{2} | vedclass

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(B) Given $a, b, c$ are in $A.P.,$ let $a = b - d$ and $c = b + d.$ Since $a+b+c=1,$ we have $(b-d) + b + (b+d) = 1,$ which implies $3b = 1$ or $b = \

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$9$

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(B) Given $a, b, c$ are in $A.P.,$ let $a = b - d$ and $c = b + d.$ Since $a+b+c=1,$ we have $(b-d) + b + (b+d) = 1,$ which implies $3b = 1$ or $b = \frac{1}{3}.$ Given $a^{2}, 2b^{2}, c^{2}$ are in $G.P.,$ we have $(2b^{2})^{2} = a^{2}c^{2},$ so $4b^{4} = (ac)^{2}.$ Substituting $a = b-d$ and $c = b+d,$ we get $4b^{4} = ((b-d)(b+d))^{2} = (b^{2}-d^{2})^{2}.$ Taking the square root, $b^{2}-d^{2} = \pm 2b^{2}.$ Case $1: b^{2}-d^{2} = 2b^{2} \Rightarrow d^{2} = -b^{2}$ (not possible for real $d$). Case $2: b^{2}-d^{2} = -2b^{2} \Rightarrow d^{2} = 3b^{2} = 3(\frac{1}{9}) = \frac{1}{3}.$ Since $a Now, $a^{2}+b^{2}+c^{2} = (b-d)^{2} + b^{2} + (b+d)^{2} = 3b^{2} + 2d^{2}.$ Substituting values: $3(\frac{1}{9}) + 2(\frac{1}{3}) = \frac{1}{3} + \frac{2}{3} = 1.$ Therefore, $9(a^{2}+b^{2}+c^{2}) = 9(1) = 9.$

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

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