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December 22, 2014

December 22, 2014

Posted by **Emily** on Friday, November 19, 2010 at 9:24pm.

- Math -
**Reiny**, Friday, November 19, 2010 at 9:55pmUse 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.

- Math -
**Bosnian**, Saturday, November 20, 2010 at 3:08amThe 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"

- Math -
**Bosnian**, Saturday, November 20, 2010 at 10:34ama1=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

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