Posted by Emily on Friday, November 19, 2010 at 9:24pm.
Use your formulas
a+9d = 34 (#1)
(20/2)[2a + 19d] = 710
2a + 19d = 71 (#2)
solve the two equations.
I would use #1 as a = 34-9d and sub into #2
Let me know what you got.
The sum of members of a arithmetic progression is:
Sn=(n/2)*[2a1+(n-1)*d]
a1-first number in arithmetic progression
d-common difference of successive members
n-numbers of members
In this case:
n=20 , (n/2)=10 , n-1=19
a1=a10-9*d
a1=34-9*d
Sn=(n/2)*[2a1+(n-1)*d]
=10*[2*(34-9*d)+19*d]
=10*(68-18d+19d)=10*(68+d)=680+10d
Sn=S20
S20=710
710=680+10d
10d=710-680=30
10d=30 Divided with 10
d=30/10
d=3
a1=a10-9d
a1=34-9*3
a1=34-27
a1=7
a25=a1+(25-1)*d
a25=7+24*3
a25=7+72
a25=79
So members of that Arithmetic progression is:
7,10,13,16,19,22,25,28,31,34,37,40,43,46,49,52,55,58,61,64,67,70,73,76,79
For more information about Arithmetic progression go to wikipedia and type"Arithmetic progression"
a1=34-9*d
Becouse nth member in arithmetic progression is:
an=a1+(n-1)*d
a10=a1+(10-1)*d
a10=a1+9*d
a10-9*d=a1
a
a1=a10-9*d
a1=34-9*d
an=a1+(n-1)*d
a25=a1+(25-1)*d
a25=a1+24*d
a25=7+24*3
a25=7+72=79