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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. `(2)/(9)`D. `(36)/(81)` |
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Answer» Correct Answer - A Let `S =` ………… `S = (4)/(19)+(44)/(19^(2))+(444)/(19^(3))+………..oo` `(S)/(19)=(4)/(19^(2))+(44)/(19^(3))+………oo` Subtracting, we get `S. (18)/(19) = (4)/(19)+(40)/(19^(2))+(400)/(1^(3))+` …………. `=(4)/(19)[1+(10)/(9)+((10)/(9))^(2)+..........]` `= (4)/(19)[(1)/(1-(10)/(19))] = (4)/(9) implies S = (76)/(162) = (38)/(81)` |
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