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

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______. [2026]

Smart Achievers · Accepted Answer

(9)

$a = b - d, c = b + d \Rightarrow b = \frac{1}{3}$

$\Rightarrow 4 b^{4} = a^{2} c^{2}$

$\Rightarrow 4 b^{4} = {[(b - d) (b + d)]}^{2}$

$\Rightarrow \frac{4}{81} = {(\frac{1}{9} - d^{2})}^{2}$

$\Rightarrow (\frac{1}{9} - d^{2}) = \pm \frac{2}{9}$

$d^{2} = \frac{1}{3} \Rightarrow d = \pm \frac{1}{\sqrt{3}} (as a < b < c)$

$∴$ $9 (a^{2} + b^{2} + c^{2})$

$= 9 [{(\frac{1}{3} - \frac{1}{\sqrt{3}})}^{2} + {(\frac{1}{3})}^{2} + {(\frac{1}{3} + \frac{1}{\sqrt{3}})}^{2}]$

$= 9 [\frac{1}{3} + \frac{2}{3}] = 3 + 6 = 9$

Solution

Q.
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______. [2026]

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Solution

Q. Suppose a,b,c are in A.P. and a2, 2b2, c2 are in G.P. If a<b<c and a+b+c=1 then 9(a2+b2+c2) is equal to______. [2026]

Ans. (9) a=b-d, c=b+d ⇒ b=13 ⇒4b4=a2c2 ⇒4b4=[(b-d)(b+d)]2 ⇒481=(19-d2)2 ⇒(19-d2)=±29 d2=13 ⇒ d=±13 (as a<b<c) ∴ 9(a2+b2+c2) =9[(13-13)2+(13)2+(13+13)2] =9[13+23]=3+6=9

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Q.
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______. [2026]