hey somebody please answer this question....
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1 + a + a² + a³ +.......∞ =x
this series are in GP
so,
x = 1/( 1- a )
1- a = 1/x
a = 1 -1/x
a= (x -1)/x
1 + b + b² + b³ +......∞ = y
y = 1/( 1- b)
1- b = 1/y
b = (y -1)/y -------(2)
again,
1 + ab + a²b² + a³b³ +........∞ =???
this series are also in GP
so, Sn = 1/( 1- ab)
= 1/{ 1- (x-1)(y-1)/xy }
=xy/( xy -xy +x +y -1)
=xy/( x + y -1)
option ( A) is correct
this series are in GP
so,
x = 1/( 1- a )
1- a = 1/x
a = 1 -1/x
a= (x -1)/x
1 + b + b² + b³ +......∞ = y
y = 1/( 1- b)
1- b = 1/y
b = (y -1)/y -------(2)
again,
1 + ab + a²b² + a³b³ +........∞ =???
this series are also in GP
so, Sn = 1/( 1- ab)
= 1/{ 1- (x-1)(y-1)/xy }
=xy/( xy -xy +x +y -1)
=xy/( x + y -1)
option ( A) is correct
mysticd:
the line below the sn =
sorry, sir but i am not reaching , can you explain . am i wrong ???
sn= 1/{1 - 1/[(x-1)(y-1)/xy] }
okay
but a = (x -1)/x not a = 1/(x -1)/x
so, why we use 1/( 1- ab) = 1/[ 1-{1/(x -1)(y-1)/xy}]
sry , u r correct
here only sn = 1/{ 1 -(x -1)(y-1)/xy }
no sorry , sometime its happen
:)
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