Given Data:

Series: \(40−10+2.5−0.625+...\)

The first term of series is: a=40

The second term of series is: \(a_2=-10\)

The third term of series is: \(a_3=2.5\)

For the first and second term,

The common ratio of the series is,

\(r=\frac{a_2}{a}\)

Substitute the values in the above equation.

\(r=\frac{-10}{40}\)

\(=-0.25\)

For the second and third term

The common ratio of the series is,

\(r=\frac{a_3}{a_2}\)

Substitute the values in the above equation.

\(r=\frac{2.5}{-10}\)

\(=-0.25\)

So, the given series is a geometric series.

The sum of the first six terms of the geometric series is,

\(S_6=a\frac{(1-r^6)}{(1-r)}\)

Substitute the values in the above equation.

\(S_6=(40)\frac{(1-(-0.25)^6)}{(1-(-0.25))}\)

\(=40\frac{(0.9998)}{1.25}\)

\(=31.99\)

Thus, the sum of the first six terms of the given geometric series is 31.99.

Series: \(40−10+2.5−0.625+...\)

The first term of series is: a=40

The second term of series is: \(a_2=-10\)

The third term of series is: \(a_3=2.5\)

For the first and second term,

The common ratio of the series is,

\(r=\frac{a_2}{a}\)

Substitute the values in the above equation.

\(r=\frac{-10}{40}\)

\(=-0.25\)

For the second and third term

The common ratio of the series is,

\(r=\frac{a_3}{a_2}\)

Substitute the values in the above equation.

\(r=\frac{2.5}{-10}\)

\(=-0.25\)

So, the given series is a geometric series.

The sum of the first six terms of the geometric series is,

\(S_6=a\frac{(1-r^6)}{(1-r)}\)

Substitute the values in the above equation.

\(S_6=(40)\frac{(1-(-0.25)^6)}{(1-(-0.25))}\)

\(=40\frac{(0.9998)}{1.25}\)

\(=31.99\)

Thus, the sum of the first six terms of the given geometric series is 31.99.