Verified Answer :

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\(a_n=5 \ for \ n=1 \\ a_n=-10 \ for \ n=2 \\ a_n=20 \ for \ n=3\)

You can see it's a geometric sequence because the ratio is constant (\(\frac{a_3}{a_2}=\frac{a_2}{a_1}\)).
The nth term of a geometric sequence:
\(a_n=a_1 \times r^{n-1}\)

\(a_1=5 \\ r=\frac{a_2}{a_1}=\frac{-10}{5}=-2 \\ \\ a_n=5(-2)^{n-1}\)

It has to be the sum for term 4 through term 15, so n=4 and the number above the sigma is 15.

Your answer:
\(\sum^{15}_{n=4} 5(-2)^{n-1}\)

Answer:

the summation of 4 times negative 3 to the n minus 1 power, from n equals 4 to 15.

Step-by-step explanation:

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