SOLUTION
The following sequence is a geometric series and we have been provided with the formula
[tex]S_n=\frac{a_1-a^{}_1r^n}{1-r}[/tex]Here a1 is the first term = 40,
r is the common ratio = -0.4 (to get r, divide the second term by the first term)
n = number of terms = 9. Now let's solve
[tex]\begin{gathered} S_n=\frac{a_1-a^{}_1r^n}{1-r} \\ \\ S_9=\frac{40_{}-40\times(-0.4)^9}{1-(-0.4)} \\ \\ S_9=\frac{40_{}-(-0.0105)^{}}{1+0.4} \\ \\ S_9=\frac{40_{}+0.0105^{}}{1+0.4} \\ \\ S_9=\frac{40.0105^{}}{1.4} \\ \\ S_9=28.5789 \end{gathered}[/tex]The sum to the nearest hundredth becomes = 28.58
