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Write the sum using summation notation, assuming the suggested pattern continues. -10 - 2 + 6 + 14 + ... + 110
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a=-10 d=-2-(-10)=-2+10=8 \[t _{n}=a+(n-1)d\] 110=-10+(n-1)*8 110+10=(n-1)*8 120/8=n-1 15+1=n or~n=16 \[t _{n}=-10+(n-1)8=-10+8n-8=-18+8n=\sum_{n=1}^{16}\left( -18+8n \right)\]
last one is Sum but not tn
\[S=\sum_{n=1}^{16}\left( -18+8n \right)\]
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