A.G.P

Asked by ravikiran | 11th Jul, 2009, 06:56: PM

Expert Answer:

assuming a  and b  tobe less than 1 sumof infinite GP

x=1+a+a2+a3....................    and y=1+b+b2+b3..........................  is

x = 1/(1-a) ....(i)

  y =1/(1-b)....(ii)

 

simplifying (i) we get   1-a =1/x and a = (x-1)/x 

Similarly (ii) will give b = (y-1)/y 

Sumof 1+ab+(ab)2+(ab)3...................................   will be

= 1/ (1-ab) 
again ab<1

Substituting the values of a and b we get 1/ (1-ab)  = xy/(x+y-1

 

Answered by  | 13th Jul, 2009, 04:01: PM

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