Question

If a, b and c be three distinct numbers in G.P. and a + b + c = xb then x can not be (A) –2 (B) –3
(C) 4 (D) 2
[JEE Main 2019]

Answer 1

answer : option (D) 2

given, a , b and c be three distinct numbers in G.P and a + b + c = xb

if a , b and c are in geometric progression.

b/a = c/b .........(1)

now a + b + c = xb

⇒x = (a + b + c)/b = a/b + 1 + c/b

⇒x = a/b + 1 + 1/(c/b)

from equation (1) we get,

= a/b + 1 + 1/(a/b)

let (a/b) = r

so, x = r + 1 + 1/r

⇒xr = r² + r + 1

⇒r² + (1 - x)r + 1 = 0

for real value of r, D = (1 - x)² - 4 ≥ 0

= 1 + x² - 2x - 4 ≥ 0

= x² - 2x - 3 ≥ 0

= (x - 3)(x + 1) ≥ 0

so, x ≥ 3 or x ≤ -1 and hence x ≠ 2

hence option (D ) is correct choice.

also read similar questions : if three distinct numbers a,b,c, are in HP, and a^2,b^2,c^2 are in AP then,

a)a+c=b

b)a+b+c=0

c)a+b=c

d)b+c=a

https://brainly.in/question/6275797

if a,b,c,d are in G.P., show that (b-c)^2+(c-a)^2+(d-b)^2=(a-d)^2

https://brainly.in/question/2119448

Answer 2

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

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

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

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

$also,$

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

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

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

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

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

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

$Then,$

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

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

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

$\therefore \: r < 0$

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

$= > x + 1 < 0$

$= > x < - 1$

$\boxed{if\:r>0\:and\:r≠1}$

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

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

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

$=>r-3>0$

$=>r>3$

$∴ x∈(−∞,−1)∪(3,∞)$

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