Dividing a term with the previous one gives the result #-5#. This means that to get the next term, we multiply by #-5#. The sequence also starts at #-2#, which is quite unconventional to express in terms of #-5#. Therefore, since the problem doesn't restrict us in any way, we can say that the equation is
#-2 * (-5)^(n-1)#.
This is because, like I mentioned above, to get to the next term we need to multiply by #-5#. Therefore, to get to the nth term, we need to multiply by #(-5)^(n-1)#.
As for why the exponent is #n-1# and not just #n#, this is because for #n= 1#, we are talking about the first term, which doesn't have anything to do with #-5#. Because of this, it's convenient to "cover" the first term with our equation by utilizing the fact that
#a^0 = 1# for any real nonzero #a#.
In other words, in order for the equation to include the case of the first number #-2# (when #n=1#) we need an expression that multiplies #-2# with not #n#, but #n-1# instances of #-5#, otherwise the first term would be #10# (and that's our second term, and so on).
