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`0.9 + 0.99 + 0.999 + ……. ` up to `51` terms `= 51 - (1)/(p)(1-(1))/(10^(q))` where `p, q in N` then find the value `(p + q)/(15)`. |
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Answer» Correct Answer - 4 `S = 0.9 + 0.99 + 0.999 +`…….up to `51` terms `= (9)/(10) + (99)/(100) + (999)/(1000) +`……. up to `51` terms `= 1 - (1)/(10) + 1 -(1)/(10^(2)) + 1 - (1)/(10^(3)) +` …………`+ 1 - (1)/(10^(51))` `= 51 - ((1)/(10) + (1)/(10^(2)) + (1)/(10^(3)) +....+(1)/(10^(51)))` `51-((1)/(10)(1-(1)/(10^(51))))/(1-(1)/(10))=51-(1)/(9)(1-(1)/(10^(51)))` `:. p + q = 60` |
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