find the sum of the first 15 terms of an arithmetic series whose nineth terms is 4o and nineteenth term is 60
In an arithmetic series:
an = a + ( n - 1 ) d
where
a = the initial term
an = the nth term
d = the common difference of successive members
a9 = a + 8 d = 40
a19 = a + 18 d = 60
Now you must solve system of two equations:
a + 8 d = 40
a + 18 d = 60
The solution is:
a = 24 , d = 2
Sum of the first n terms:
Sn = n [ 2 a + ( n - 1) d ] / 2
In this case:
n = 15 , a = 24 , d = 2
S15 = 15 ∙ [ 2 ∙ 24 + ( 15 - 1 ) ∙ 2 ] / 2 =
15 ∙ ( 48 + 14 ∙ 2) / 2 = 15 ∙ ( 48 + 28 ) / 2
= 15 ∙ 76 / 2 = 1140 / 2 = 570