I can't figure out how to find the n on top of the sigma. I have the rest of the equation but I just don't know what to do. I'll use "E" for sigma.

Question

nEi=1 (10-3i)=-28

Can someone please show me how to do this problem?

Answer 1

n
∑ (10-3i) = -28
i=1

You know that

n
∑ 10 = 10n
i=1

You know that

n
∑ i = n(n+1)/2
i=1

so

n
∑ -3i = -3n(n+1)/2
i=1

Now just add them up:

10n - 3n(n+1)/2 = -28
n = 8