JEE Main 20249 Apr, Morning shiftMathematicsMediumSequence and Series179× since 2002If the sum of the series 11⋅(1+d)+1(1+d)(1+2 d)+…+1(1+9 d)(1+10 d)\frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}1⋅(1+d)1+(1+d)(1+2 d)1+…+(1+9 d)(1+10 d)1 is equal to 5, then 50 d50 \mathrm{~d}50 d is equal to :AOption A. 5BOption B. 10COption C. 15DOption D. 20CheckSkip · show solution
