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Question Number 157781 by amin96 last updated on 27/Oct/21

Answered by mr W last updated on 27/Oct/21

$$100A=(\frac{1}{1}-\frac{1}{101})+(\frac{1}{2}-\frac{1}{102})+...+(\frac{1}{25}-\frac{1}{125})$$$$25B=(\frac{1}{1}-\frac{1}{26})+(\frac{1}{2}-\frac{1}{27})+...+(\frac{1}{100}-\frac{1}{125})$$$$25B-100A=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{125})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{125})=0$$$$25B=100A$$$$\frac{A}{B}=\frac{25}{100}=\frac{1}{4}$$

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