If a,b,c, are in G.P. and x,y respectively arithmetic means between a,b and b,c then prove that
$\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$

Open in App

Solution

$\frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c} = \frac{2 (a + c + 2 b)}{a b + a c + b^{2} + b c}$
put $b^{2} = a c$
$= \frac{2 (a + c + 2 b)}{a b + 2 b^{2} + b c} = \frac{2 (a + c + 2 b)}{b (a + c + 2 b)} = \frac{2}{b}$

Suggest Corrections

Similar questions

If a, b,and c are in G.P. and x, y, respectively, be arithmetic means between a, b and b, c, then the value of $\frac{a}{x} + \frac{c}{y}$ is

Q. If

a, b, c

are in

G . P

and

x, y

, respectively be arithmetic means between

a, b

and

b, c

then the value of

\frac{a}{x} + \frac{c}{y}

Q. Let

a, b, c

be in G.P. If

x, y

are arithmetic means between

a, b

and

b, c

respectively, then which of the following is/are correct?

Q. Let

a, b, c

be in G.P. If

x, y

are arithmetic means between

a, b

and

b, c

respectively, then which of the following is/are correct?

Find the sum of first n terms of the series $3 + 7 + 13 + 21 + 31 + . . . .$

If a b and c are in GP and x,y are the arithmetic means of a,b and b,c respectively. prove that $\frac{a}{x} + \frac{c}{y} = 2 a n d \frac{1}{x} + \frac{1}{y} = \frac{2}{b}$ .

Join BYJU'S Learning Program

If a,b,c, are in G.P. and x,y respectively arithmetic means between a,b and b,c then prove that1x+1y=2b

1x+1y=2a+b+2b+c=2(a+c+2b)ab+ac+b2+bcput b2=ac =2(a+c+2b)ab+2b2+bc=2(a+c+2b)b(a+c+2b)=2b

If a,b,c, are in G.P. and x,y respectively arithmetic means between a,b and b,c then prove that
$\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$

$\frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c} = \frac{2 (a + c + 2 b)}{a b + a c + b^{2} + b c}$
put $b^{2} = a c$
$= \frac{2 (a + c + 2 b)}{a b + 2 b^{2} + b c} = \frac{2 (a + c + 2 b)}{b (a + c + 2 b)} = \frac{2}{b}$