Question

solve it very hard Question
If a,b,c are G.P then a+b+c=xb. Then x≠.​

Answer 1

Step-by-step explanation:

General team of a GP -

$tn = {af}^{n} - 1$

_ When ein

a => First team

r => Common ratio

From the concept have teams be

Let the three teams be

b/r, b, be => in GP

Given: a + br = x b

$= \frac{b}{r} + b \: + br = xb$

$= \frac{1}{r} + 1 + r = x$

$= x - 1 = r + \frac{1}{r}$

Hope it's help you....

Answer 2

$\fcolorbox{pink}{white} {\tt{answer}}$

$a,b,c are \: in \: GP$

$Let \: 'r' \: be \: the \: common \: ratio \: then,$

$\frac{b}{a} = r \: \: = > a = \frac{b}{r}$

$also,$

$\frac{c}{b} = r \: \: \: = >c = br$

$\therefore \: a + b + c = \frac{b}{r} + r + br = b( \frac{1 + r + {r}^{2} }{r} )$

$if \: a + b + c = xb$

$then \: ( \frac{1 + r + {r}^{2} }{r} ) = x$

$\boxed{if \: r < 0 \: and \: r ≠ - 1}$

$( \therefore \: a,b,c \: are \: 3 \: distinct \: no.)$

$Then,$

$\frac{1 + r + {r}^{2} }{r} = x$

$= > \frac{1 + r + {r}^{2} }{r} + 1 = x + 1$

$= > \frac{(1 + r {)}^{2} }{r} = x$

$\therefore \: r < 0$

$There \: by, \: \frac{(1 + r {)}^{2} }{r} < 0( \therefore \: (1 + r {)}^{2} > 0)$

$= > x + 1 < 0$

$= > x < - 1$

$\boxed{if\:r>0\:and\:r≠1}$

$then, \:\frac{1 + r + {r}^{2} }{r}=x \: \: =>\frac{1 + r + {r}^{2} }{r} - 3=x-3$

$= > \frac{(r-1{)}^{2} }{r}=x-3$

$Again\:\frac{(r-1{)}^{2} }{r}>0(\therefore \:r>0\: and\:(1+r)^{2}<0)$

$=>r-3>0$

$=>r>3$

$∴ x∈(−∞,−1)∪(3,∞)$

$\small{\boxed{\pink{Hence,\:x\:can\: not\:be\:2}}}$