Math Arithmetic Series?
The question sounds like this:
"The sum of the first twenty terms of an arithmetic series is 540 and the sum of the first thirty terms of the series is 1260.
Find the first term and the common difference of the series"
I was wondering how you guys would work this out? I have tried for a long time now, and I cant quite get my head around it. I have reduced them to:
S20 = 540
540 = 20u1 + 190d
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S30 = 1260
1260 = 30u1 + 435d
I have attempted to subtract the value of S20 with S30, but I got stuck form there! 
Could you guys please help me out a little?
Thank you very much on-forehand!
Comments
540/20 = 27, the average of the first 20 terms
1260/30 = 42, the average of the first 30 terms
d = (42-27)/5 = 3, the common difference
[a(1)+a(1)+(20-1)*3]*20/2 = 540
a(1) = -3/2
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Get it done in the simplest way is what you always should try to learn.
Sn=n/2*{2a+(n-1)d} where S is the sum , n is the number of terms , a is the first term, d is the difference
S20= 20/2*{2a+(20-1)d}=540 so ==> 540=10(2a+19d) divide by 10
so 2a+19d=54 ...........(1)
S30=30/2 *{2a+(30-1)d}=1260 so 15(2a+29d)=1260 divide by 15
so 2a+29d=84 .......(2) subtract (1) from (2)
so 10d=30 so d=30/10=3
from (1) 2a+3(19)=54 , 2a+57=54 so a=54-57=-3
so the first term=--3 and the common difference =3
thanks (fawzy-maths teacher-egypt)tel 0020101238946
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20u1 + 190d = 540
divide by 10
2u1 + 19d = 54-----------------------eqn(1)
1260 = 30u1 + 435d
divide by 15
2u1 + 29d = 84----------------------eqn(2)
subtract (1) from (2)
10d = 30
d = 3
substitute d value in eqn (1)
2u1 + 57 = 54
2u1 = -3
u1 = -1.5
first term = -1.5
common difference = 3