Question 945097
Both problems require the formula to find the nth term of an arithmetic sequence:
Sn = A1 + (n-1)d
where
Sn is "sum of n terms"
A1 is the first term
n is the number of terms
d is the distance
.
Problem 1: a1= 4, d = 7, Sn= 228 
Sn = A1 + (n-1)d
substitute the values given from the problem:
228 = 4 + (n-1)7
228 = 4 + 7n-7
228 = 7n-3
231 = 7n
33 = n
.
Problem 2: (−2) + (−12) + (−22) + (−32)..., Sn= −224 
distance (d) is:
-12 - (-2) = -12 + 2 = -10
.
Sn = A1 + (n-1)d
substitute the values given from the problem:
-224 = -4 + (n-1)-10
-224 = -4 + -10n+10
-224 = -10n+6
-230 = -10n
23 = n