THE FIRST AND LAST TERMS OF AN ARITHMETIC SERIES IS 5 AND 61 RESPECTIVELY WHILE THE SUM OF ALL THE TERMS IS 957. DETERMINE THE NUMBER OF TERMS IN THE SERIES.

Answer 1

If there are n terms,

n/2 (5+61) = 957
n = 29

Answer 2

a = 5
last term = 61

sum(n) = (n/2)(first + last)
957 = (n/2)(5 + 61)
1914 = 66n
n =29