Find the first three terms of an arithmetic series in which the fifteenth term is –7 and the sum of the first fifteen terms is 210.
We have
-7 = a1 + (14)d → 14d = -7 - a1 (1)
210 = (15/2) (2a1 + 14(d) ) (2)
Sub (1) into (2)
210 = (15/2) ( 2a1 -7 - a1 )
210 ( 2 / 15) = a1 - 7
28 = a1 - 7
a1 = 35
And
14d = -7 - 35 → 14d = - 42 → d = -42/14 = -3
a1 = 35 a2 = 32 a3 = 29
