Math
posted by Emily on .
The 10th term of an arithmetic series is 34, and the sum of the first 20 terms is 710. Determine the 25th term.

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 = 349d and sub into #2
Let me know what you got. 
The sum of members of a arithmetic progression is:
Sn=(n/2)*[2a1+(n1)*d]
a1first number in arithmetic progression
dcommon difference of successive members
nnumbers of members
In this case:
n=20 , (n/2)=10 , n1=19
a1=a109*d
a1=349*d
Sn=(n/2)*[2a1+(n1)*d]
=10*[2*(349*d)+19*d]
=10*(6818d+19d)=10*(68+d)=680+10d
Sn=S20
S20=710
710=680+10d
10d=710680=30
10d=30 Divided with 10
d=30/10
d=3
a1=a109d
a1=349*3
a1=3427
a1=7
a25=a1+(251)*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=349*d
Becouse nth member in arithmetic progression is:
an=a1+(n1)*d
a10=a1+(101)*d
a10=a1+9*d
a109*d=a1
a
a1=a109*d
a1=349*d
an=a1+(n1)*d
a25=a1+(251)*d
a25=a1+24*d
a25=7+24*3
a25=7+72=79