a = 9
term(n) = a + (n-1)d = 105
(n-1)d = 105 - a = 105-9 = 96
sum(n) = (n/2)(2a + (n-1)d) = 741
(n/2)(18 + 96) = 741
n/2 = 6.5
n = 13
(n-1)d = 96
12d = 96
d = 8
the terms are 9, 17, 25, ..
What are the first three terms of an arithmetic series in which a1=9,an=105,and Sn=741?
1 answer