106as l r l <1 so S = a/(1-r)
So, x = 1/(1-a) and y = 1/(1-b) OR a = 1-1/x and b=1-1/y
For S of the reqd GP, r = ab .. and as lal,lbl<1 so, labl<1
Hence S = 1/(1-ab) put the values of a and b obtained above
x=1+a+a2+a3+...to infinity.(|a|<1)
y=1+b+b2+...to infinity.(|b|<1)
Prove that 1+ab+a2b2+a3b3+...to infinity=xy/x+y-1
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4 Answers
106
106Sum of n terms of a GP = a(1-rn)/(1-r)
Now if lrl < 1 then the sum converges to a particular value. as lrl<1 so lim(n→∞)rn = 0
So, the sum can be re-written as a/(1-r)
In 1st GP, a=1 and r=a
In 2nd GP, a=1 and r=b