1st term --- a
15th term = a + 14d = 9a
14d = 8a
d = 4a/7
sum(35) = (35/2)(2a + 34d)
= (35/2)(2a + 34(4a/7)
= (35/2)(150a/7)
= 375a
Find the sum of 35 terms of an arithmetic series of which the first term is "a" and the fifteenth term is "9a."
1 answer