Compare a geometric training total with an arithmetic total

$u_{30}=50(1.1)^{29}\approx793.15465$, giving $793$ m to the nearest metre.

$T=50[(1.1)^{30}-1]/(1.1-1)\approx8224.70113$, giving $8225$ m to the nearest metre.

For B, $S_n=n[2(50)+(n-1)5]/2=n(5n+95)/2$. Using the source’s rounded target $8225$ gives $n^2+19n-3290>0$, with positive cutoff about $48.64$. Thus the least positive integer candidate is $49$. Using A’s unrounded total gives the same least integer.

The source calculator table is:

These totals increase, and $8040<T<8330$, so the least number of days is $49$. The source’s intermediate decimal cutoff is not a fractional permissible day count; round to the next integer satisfying the strict inequality.