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Question Number 27197 by mondodotto@gmail.com last updated on 03/Jan/18

Commented by prakash jain last updated on 03/Jan/18

I think there is typo (3/5),((10)/8),((16)/(18)),((36)/(34)),((68)/(70)),((140)/(138)),((276)/(268)) ((Numerator of n^(tb) term)/(Deniminator of n^(th) term)) =((2×Deniminator of n−1^(th) term)/(Numerator of n−1^(tb) term+Deniminator of n−1^(th) term))

$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{typo} \\ $$$$\frac{\mathrm{3}}{\mathrm{5}},\frac{\mathrm{10}}{\mathrm{8}},\frac{\mathrm{16}}{\mathrm{18}},\frac{\mathrm{36}}{\mathrm{34}},\frac{\mathrm{68}}{\mathrm{70}},\frac{\mathrm{140}}{\mathrm{138}},\frac{\mathrm{276}}{\mathrm{268}} \\ $$$$\frac{\mathrm{Numerator}\:\mathrm{of}\:{n}^{{tb}} \:\mathrm{term}}{\mathrm{Deniminator}\:\mathrm{of}\:{n}^{{th}} \:\mathrm{term}} \\ $$$$=\frac{\mathrm{2}×\mathrm{Deniminator}\:\mathrm{of}\:{n}−\mathrm{1}^{{th}} \:\mathrm{term}}{\mathrm{Numerator}\:\mathrm{of}\:{n}−\mathrm{1}^{{tb}} \:\mathrm{term}+\mathrm{Deniminator}\:\mathrm{of}\:{n}−\mathrm{1}^{{th}} \:\mathrm{term}} \\ $$

Answered by prakash jain last updated on 05/Jan/18

Also see question 27300

$$\mathrm{Also}\:\mathrm{see}\:\mathrm{question}\:\mathrm{27300} \\ $$

Commented by mondodotto@gmail.com last updated on 05/Jan/18

ok thanx

$$\mathrm{ok}\:\mathrm{thanx} \\ $$

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