JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 68

$$S=\sum_{k=1}^{10}k\,11^{k-1}10^{10-k}=10^9\sum_{k=1}^{10}k\,r^{k-1},\qquad r=\tfrac{11}{10}.$$
Using $\sum_{k=1}^nkr^{k-1}=\dfrac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}$ with $n=10$:
$$\frac{1-11r^{10}+10r^{10}\cdot\frac{11}{10}}{0.01}=\frac{1}{0.01}=100.$$
So $k=100$.