An arithmetic progression can be written a, a+d, a+2d, etc., and its sum to n terms is a+a+d+a+2d+a+3d+...+a+(n-1)d=na+d∑i[0,n-1]. (a is the first term and d the common difference. The square brackets show the lower and upper limits for i in the sum, represented by sigma.) So we have pairs of terms (a+a+(n-1)d)+(a+d+a+(n-2)d)+(a+2d+a+(n-3)d)+... In the given series we have the sum -2+(-5)+(-8)+(-11), that is, the first 4 terms. We can write this: (-2-11)+(-5-8)=-13-13=-26. So by arranging the terms in pairs we have 2 pairs each with the same sum: a+a+(n-1)d, that is, 2a+(n-1)d times n/2. If we put a=-2, d=-3 and n=4, we get (-4-9)(4/2)=-26.
We can now find an expression for the sum S of the series for a general n.
S=(2a+(n-1)d)(n/2)=an+dn(n-1)/2. Applying it when n=8 we get S=(-4-21)(8/2)=-100, or -16+(-3)(56/2)=-16-84=-100.
Let’s check: series is -2, -5, -8, -11, -14, -17, -20, -23=(-2-23)+(-5-20)+(-8-17)+(-11-14)=-25×4=-100.