If {1+3+5+. . . .+ {n} \text{ terms }}/{2+4+6+. . . . . . 50 \text{ terms }}=\frac{2}{51}, then the value of 'n'

Sequence and Series

Question : 73 of 86

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\;{1+3+5+. . . .+ {n} \;\text " terms "\;}/{2+4+6+. . . . . . 50 \;\text " terms "\;}=\;{2}/{51},

then the value of 'n'

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