As you noted, on the $30$-th day the lady walked a distance of $2950$ metres.
You were not told the full story. The lady lives at A, and her lover lives at B, which is $3000$ metres from A. On Day $1$, the lady walked $50$ metres, and her lover carried her the remaining $2950$ metres to B. On Day $2$, the lady walked $150$ metres, and was carried the remaining $2850$ metres to B. On Day $3$, the lady walked $250$ metres, and was carried $2750$ metres. And so on. On Day $30$, the lady walked $2950$ metres and was carried $50$ metres.
So the distances she walked went $50, 150, 250, \dots, 2750, 2850, 2950$, for a total of
$$50+150+250+ \cdots +2750+ +2850+2950.$$
We could simply use a calculator to add up. But there is a better way.
The distances she was carried went $2950, 2850, 2750, \dots, 250, 150, 50$. The total distance she was carried (we are adding up backwards) is therefore
$$50+150+250+ \cdots +2750+ 2850+2950,$$
which is the same as the total distance she walked.
Every day, the distance walked plus the distance carried is $3000$, so in $30$ days, total distance walked plus total distance carried is $(30)(3000)$. But total distance walked is equal to the total distance carried, so the total distance walked is
Comment: Suppose that she walks a distance of $a$ the first day, and adds distance $d$ to her workout every day. Suppose also that she does this a total of $n$ days. Then on the last day she walks $a+(n-1)d$. Same story, except lady and lover live $2a+(n-1)d$ apart. The same argument shows that
Alternately, if the distance walked on Day $1$ is $a_1$, and increases by the same amount each day, reaching distance $a_n$ on Day $n$, then the total distance walked is