Question

what is the sum of the first 11 terms of the arithmetic series in which
asub one = -12 and d=5

Answer 1

so a = -12 , d = 5

sum(n) = (n/2)(2a + (n-1)d )
= (11/2)(-24 + 10(5))
= (11/2)(26)
= 143

OR

sum(n) = (n/2)(first + last)
= (11/2)(-12 + 38)
= 143

Answer 2

An = A1 + (n-1)d

average = (A1 + An)/2
sum = average * number of terms
= n (A1 +An)/2

= (n/2) [ 2 A1 + (n-1)d ]

here
= (11/2) [ -24 + (10)5 ]

= (11/2) (26) = 11 * 13 = 143