Bài giải :

$D=\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}$

$D=\frac{15}{4}\left( \frac{4}{90.94}+\frac{4}{94.98}+\frac{4}{98.102}+...+\frac{4}{146.150} \right)$

$D=\frac{15}{4}\left( \frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150} \right)$

$D=\frac{15}{4}\left( \frac{1}{90}-\frac{1}{150} \right)$

$D=\frac{1}{60}$

Vậy D<1.