Question 881058
the fifteenth term in an arithmetic sequence is 43 and the sum of the first 15 terms of the series is 120. determine the first three terms of the sequence.
T15= a+14d=43  (1)
S15= 15(2a+14d)/2 =120   (2)
15(a+a+14d)/2 =120  by substituting a+14d=43 in (2)
15(a+43)/2=120
a+43=120x2/15
a+43=16
a=16-43 =-27
a+14d=43
-27 +14d = 43
14d=43+27
14d = 70
d=70/14 =5
Answer  first terms= -27
        Second term = -27+5 =-22
        third term  = -27+2x5 =-27 +10= -17