If a, b, c are in AP; b, c, d are in GP and c, d, e are in HP, then a, c, e are in


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Question Number 7524 by Little last updated on 02/Sep/16

$If a, b, c are in AP; b, c, d are in GP and$ $c, d, e are in HP, then a, c, e are in$
Commented by Rasheed Soomro last updated on 02/Sep/16

$C o m p l e t e A n s w e r$ $If a, b, c are in AP; b, c, d are in GP and$ $c, d, e are in HP, then a, c, e are in$ $- - - - - - - - - - - - - - - - - - -$ $a, b, c a r e i n A P :$ $c = a + 2 d^{'} = a + 2 (b - a) = 2 b - a [d^{'} i s c o m m o n d i f f e r e n c e]$ $b, c, d a r e i n G P :$ $d = {b r}^{2} = b {(c / b)}^{2} = c^{2} / b = \frac{{(2 b - a)}^{2}}{b} [r i s c o m m o n r a t i o]$ $c, d, e a r e i n H P \Rightarrow d = \frac{2 c e}{c + e} \Rightarrow e = \frac{d c}{2 c - d}$ $e = \frac{d c}{2 c - d} = \frac{\frac{{(2 b - a)}^{2}}{b} \times (2 b - a)}{2 (2 b - a) - \frac{{(2 b - a)}^{2}}{b}} = \frac{\frac{{(2 b - a)}^{3}}{b}}{\frac{2 b (2 b - a) - {(2 b - a)}^{2}}{b}}$ $= \frac{{(2 b - a)}^{3}}{(2 b - a) [2 b - 2 b + a]} = \frac{{(2 b - a)}^{2}}{a}$ $a, c, e$ $a, 2 b - a, \frac{{(2 b - a)}^{2}}{a}$ ${L e t}^{'} s a s s u m e t h a t t h i s i s G P$ $T e s t i n g f o r c o m m o n r a t i o$ $\frac{2 b - a}{a} \overset{?}{=} \frac{\frac{{(2 b - a)}^{2}}{a}}{2 b - a}$ $\frac{2 b - a}{a} \overset{?}{=} \frac{{(2 b - a)}^{2}}{a} \times \frac{1}{2 b - a}$ $\frac{2 b - a}{a} = \frac{2 b - a}{a}$ $H e n c e a, 2 b - a, \frac{{(2 b - a)}^{2}}{a} i s a G P, w h o s e c o m m o n r a t i o$ $i s \frac{2 b - a}{a}$ $O r a, c, e a r e i n G P$
	Answered by Rasheed Soomro last updated on 07/Sep/16

	$A n s w e r i s i n c o m m e n t .$
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