find the sum of the first 15 terms of an arithmetic series whose nineth terms is 4o and nineteenth term is 60

Answer 1

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