JEE Main 202229 Jul, Morning shiftMathematicsEasyComplex Numbers and Quadratic Equations264× since 2002If 1(20−a)(40−a)+1(40−a)(60−a)+…+1(180−a)(200−a)=1256\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}(20−a)(40−a)1+(40−a)(60−a)1+…+(180−a)(200−a)1=2561, then the maximum value of a\mathrm{a}a is :AOption A. 198BOption B. 202COption C. 212DOption D. 218CheckSkip · show solution
