Question

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

Answer 1

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.

Answer 2

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"

Answer 3

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