Find the sum of 35 terms of an arithmetic series of which the first term is "a" and the fifteenth term is "9a".
a + 14d = 9a → 8a = 14d → d = [8/14]a = (4/7)a
So
S35 = (35/2) [2a + 34 (4/7) a ]
S35 = (35/2) [ 2a + (136/7)a ]
S35 = (35/2) [ (150/7)a ]
S35 = ( 35/7)(150/2) a = (5)(75) a = 375 a
