1.

Let S=`4/19+44/(19)^2+444/(19)^3+...oo` then find the value of SA. `(38)/(81)`B. `(4)/(19)`C. `(36)/(871)`D. `(4)/(9)`

Answer» Correct Answer - A
The sum of the series ………..
`S = (4)/(19)+(44)/(19^(2))+………… oo`
`(S)/(19) = (4)/(19^(2))+(44)/(19^(3))+………….oo`
`(18S)/(19)= (4)/(19)[1+(10)/(19)+(10^(2))/(19^(2))+………..oo]`
`s = (38)/(81)`


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