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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)` |
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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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