Suppose a , b , c are in A.P. and a 2 , b 2 , c 2 are in G.P. If a b c and a + b + c = 3 2 , then the value of a is [2002]

Question

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = \frac{3}{2}$ , then the value of $a$ is [2002]

Smart Achievers · Accepted Answer

(4)

$Since a, b, c are in A.P.$

$∴$ $2 b = a + c$

$But given a + b + c = \frac{3}{2} \Rightarrow b = \frac{1}{2} and then a + c = 1$

$Also a^{2}, b^{2}, c^{2} are in G.P. \Rightarrow b^{4} = a^{2} c^{2}$

$\Rightarrow b^{2} = \pm a c \Rightarrow a c = \frac{1}{4} or - \frac{1}{4}$

$and a + c = 1 . . . (i)$

$Considering a + c = 1 and a c = \frac{1}{4}$

${(a - c)}^{2} = 1 - 1 = 0 \Rightarrow a = c but a \neq c$

$as given that a < b < c$

$∴$ $a + c = 1 and a c = - \frac{1}{4}$

$\Rightarrow {(a - c)}^{2} = 1 + 1 = 2 \Rightarrow a - c = \pm \sqrt{2}$

$but a < c \Rightarrow a - c = - \sqrt{2} . . . (i i)$

$Solving (i) and (ii), we get a = \frac{1}{2} - \frac{1}{\sqrt{2}}$

Solution

Q.
Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = \frac{3}{2}$ , then the value of $a$ is [2002]

1 $\frac{1}{2 \sqrt{2}}$

2 $\frac{1}{2 \sqrt{3}}$

3 $\frac{1}{2} - \frac{1}{\sqrt{3}}$

4 $\frac{1}{2} - \frac{1}{\sqrt{2}}$

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Solution

Q. Suppose a,b,c are in A.P. and a2,b2,c2 are in G.P. If a<b<c and a+b+c=32, then the value of a is [2002]

1 122

2 123

3 12-13