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

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Question

(d) $b = a + d,c = a + 2d$, where $d > 0$

Now ${a^2},{(a + d)^2},{(a + 2d)^2}$are in $G.P.$

$	herefore {(a + d)^4} = {a^2}{(a + 2d)^2}$

o

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) $b = a + d,c = a + 2d$, where $d > 0$ Now ${a^2},{(a + d)^2},{(a + 2d)^2}$are in $G.P.$ $ herefore {(a + d)^4} = {a^2}{(a + 2d)^2}$ or ${(a + d)^2} = \pm a(a + 2d)$ or ${a^2} + {d^2} + 2ad = \pm \,\,({a^2} + 2ad)$ Taking $(+)$ sign, $d = 0$ (not possible as $a < b < c)$ Taking $(-)$ sign, $2{a^2} + 4ad + {d^2} = 0$, $\left[ {a + b + c = \frac{3}{2},\,\,\, herefore a + d = \frac{1}{2}} ight]$ $2{a^2} + 4a\left( {\frac{1}{2} - a} ight) + {\left( {\frac{1}{2} - a} ight)^2} = 0$ or $4{a^2} - 4a - 1 = 0$ $ herefore a = \frac{1}{2} \pm \frac{1}{{\sqrt 2 }}.{ m{Here}}\,{ m{ }}d = \frac{1}{2} - a > 0.$ So, $a < \frac{1}{2}.$ Hence $a = \frac{1}{2} - \frac{1}{{\sqrt 2 }}$.

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

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