If-three-positive-numbers-a-b-c-are-in-A-P-and-1-a-2-1-b-2-1-c-2-also-in-A-P-then-1-a-b-c-2-2b-3a-c-3-b-2-ac-8-4-2c-2b-a-

Question Number 24858 by Tinkutara last updated on 27/Nov/17

If three positive numbers a, b, c are in

A . P . and \frac{1}{a^{2}}, \frac{1}{b^{2}}, \frac{1}{c^{2}} also in A . P ., then

(1) a = b = c

(2) 2 b = 3 a + c

(3) b^{2} = \frac{a c}{8}

(4) 2 c = 2 b + a

Answered by ajfour last updated on 28/Nov/17

l e t a = b - d

a n d c = b + d

\frac{2}{b^{2}} = \frac{a^{2} + c^{2}}{a^{2} c^{2}} = \frac{2 (b^{2} + d^{2})}{{(b^{2} - d^{2})}^{2}}

l e t d = k b

\Rightarrow \frac{1}{b^{2}} = \frac{(1 + k^{2}) b^{2}}{{(1 - k^{2})}^{2} b^{4}}

o r {(1 - k^{2})}^{2} = 1 + k^{2}

l e t 1 - k^{2} = t

\Rightarrow t^{2} = 2 - t

\Rightarrow {(t + \frac{1}{2})}^{2} = \frac{9}{4}

o r t = 1

\Rightarrow 1 - k^{2} = 1 \Rightarrow k = 0

s o a = b = c .

Commented by prakash jain last updated on 30/Nov/17

t = - 2

k = \sqrt{3}

b = b

a = b (1 - \sqrt{3})

c = b (1 + \sqrt{3})

That is the solution provided

by jota + .

Commented by Tinkutara last updated on 28/Nov/17

Thank you Sir!

Commented by prakash jain last updated on 30/Nov/17

What about the solution posted

below by jota + . I have put

my calculation as a comment there .

Answered by jota+ last updated on 29/Nov/17

o n e s o l u t i o n

a = 1 - \sqrt{3} b = 1 c = 1 + \sqrt{3}

Commented by prakash jain last updated on 30/Nov/17

o k .

Commented by prakash jain last updated on 30/Nov/17

\frac{1}{a^{2}} = \frac{1}{4 - 2 \sqrt{3}} = \frac{1}{2 (2 - \sqrt{3})}

= \frac{2 + \sqrt{3}}{2} = 1 + \frac{\sqrt{3}}{2}

\frac{1}{b^{2}} = 1

\frac{1}{c^{2}} = \frac{1}{4 + 2 \sqrt{3}} = \frac{1}{2 (2 + \sqrt{3})} = 1 - \frac{\sqrt{3}}{2}

Commented by Tinkutara last updated on 30/Nov/17

But my ques . says a, b, c to be + ve .