Answer
See below
Work Step by Step
We will consider the sum as the sum of geometric series:
$0.151515...=0.15+0.0015+0.000015+...\\=0.15(1+0.01+0.0001+...)\\=\sum^{\infty}_{i=1}0.15(0.01)^{i-1}$
We can notice that $a_1=0.15\\r=0.01$
Hence, the required number is: $0.151515...=\sum^{\infty}_{i=1}0.15(0.01)^{i-1}\\=\frac{0.15}{1-0.01}\\=\frac{0.15}{0.99}\\=\frac{15}{99}\\=\frac{5}{33}$
