Series: (96+93+90+\cdots)

(a=96), (d=-3), (S_n=1071)

[ \dfrac{n}{2}\bigl[2\cdot96+(n-1)(-3)\bigr]=1071 ] [ n(192-3n+3)=2142 ] [ n(195-3n)=2142 ] [ 195n-3n^2=2142 ] [ 3n^2-195n+2142=0 ] [ n^2-65n+714=0 ] [ (n-17)(n-42)=0 ]

(n=17) or (n=42). For (n=42), later terms become negative and the sum starts decreasing after a point; both can satisfy the equation, but the least positive number of terms is often taken. Checking (S_{17}):

[ S_{17}=\dfrac{17}{2}[192+16\cdot(-3)]=\dfrac{17}{2}(192-48)=\dfrac{17}{2}\cdot144=1224 ] Wait recalculate:

[ S_n=\frac{n}{2}[192-3(n-1)]=\frac{n}{2}(195-3n) ] (\frac{17}{2}(195-51)=\frac{17}{2}\cdot144=1224\neq1071)

From quadratic: (n^2-65n+714=0) Discriminant: (4225-2856=1369=37^2) (n=\frac{65\pm37}{2}=51) or (14)

Recheck equation: [ \frac{n}{2}[192-3(n-1)]=1071 ] [ n[192-3n+3]=2142 ] [ n(195-3n)=2142 ] [ 195n-3n^2=2142 ] [ 3n^2-195n+2142=0 ] [ n^2-65n+714=0 ] [ n=\frac{65\pm\sqrt{4225-2856}}{2}=\frac{65\pm37}{2} ] (n=51) or (n=14)

[ S_{14}=\frac{14}{2}[192-3\cdot13]=7(192-39)=7\cdot153=1071 ]

Yes.

Answer: 14 terms (also (n=51) mathematically satisfies)