Answer with explanation:
[tex]\sum_{i=2}^{i=7}3i-15\\\\=(3*2-15)+(3*3-15)+(3*4-15)+(3*5-15)+(3*6-15)+(3*7-15)\\\\=3*2+3*3+3*4+3*5+3*6+3*7-(15+15+15+15+15+15)\\\\=3*(1+2+3+4+5+6+7)-3-90\\\\=3*\frac{7*(1+7)}{2}-93\\\\=3*28-93\\\\=84-93\\\\=-9\\\\ \text{Sum of n terms of an A.P}=\frac{n*(a+l)}{2}[/tex]
where a, is first term and l is last term.
Option C= -9
