## anonymous one year ago Write the sum using summation notation, assuming the suggested pattern continues. -10 - 2 + 6 + 14 + ... + 110

1. anonymous

l=a+(n-1)d a=-10 l=110 d=-2-(-10)=8 110=-10+(n-1)8=-10+8n-8=8n-18 110+10=(n-1)8 120/8=n-1 n-1=15 n=15+1=16 solve $\sum_{1}^{16}\left( 8n-18 \right)$

2. mathstudent55

a1 = -10 a2 = -10 + 8 = a1 + 8 a3 = -10 + 8 + 8 = a1 + 8(2 - 1) a4 = -10 + 8 + 8 + 8 = a1 + 8(3 - 1) an = a1 + 8(n - 1) an = -10 + 8(n - 1) = -10 + 8n - 8 an = -18 + 8n an = 8n - 18 $$\Large \sum \limits_{n = 11}^{16} ( 8n-18 )$$

3. anonymous

These are my answer choices: summation of negative eighty times n from n equals zero to fifteen summation of the quantity negative ten plus eight n from n equals zero to fifteen summation of the quantity negative ten plus eight n from n equals zero to infinity summation of negative eighty times n from n equals zero to infinity

4. anonymous

@xapproachesinfinity