# Arithmetic series

OK, these questions are supposed to be really easy, but I keep geting the wrong answer and I don't understand why, could someone pleas look at my work and tell me what I am doing wrong?

Find the indicated sum of each arithmetic series

S15 for 4 + 11 + 18 + ...

Sn= n/2 [ 2a + ( n - 1 ) d ]
Sn= 15/2 [ 2(4) + (15 - 1 ) 7 ]
Sn= 7.5 [ 8 + 98 ]
Sn= 795

Find t8 for the following sequence

10 000 , -5 000 , 2 500 , -1 250 ,

So...

tn= ar ^ n-1
tn= 10 000 (-0.5) ^ n-1

t8= 10 000 (-0.5) ^ 7
= 78.125

Sometimes the answers in the back of the book are wrong, so am I wrong or are they??

HallsofIvy
Homework Helper
Aya said:
Find the indicated sum of each arithmetic series

S15 for 4 + 11 + 18 + ...

Sn= n/2 [ 2a + ( n - 1 ) d ]
Sn= 15/2 [ 2(4) + (15 - 1 ) 7 ]
Sn= 7.5 [ 8 + 98 ]
Sn= 795
Presumably you not that if an has initial value a0 and "constant difference" x, then an= a0+n(xn. Furthermore, the sum of the first n terms of an arithmetic series is $a_0+ a_n= (a_0+ x_n)n/2$. That is, the sum is n times the "average" of the first and last terms of the sequence.
In the first problem, a0= 4 and r= 11- 4= 7. The first term is 4 and the 15th term is 4+ 14(7)= 4+ 98= 102. The sum is (4+ 102)/2= (106)/2= 53.

VietDao29
Homework Helper
HallsofIvy said:
Presumably you not that if an has initial value a0 and "constant difference" x, then an= a0+n(xn. Furthermore, the sum of the first n terms of an arithmetic series is $a_0+ a_n= (a_0+ x_n)n/2$. That is, the sum is n times the "average" of the first and last terms of the sequence.
In the first problem, a0= 4 and r= 11- 4= 7. The first term is 4 and the 15th term is 4+ 14(7)= 4+ 98= 102. The sum is (4+ 102)/2= (106)/2= 53.

Uhm , the sum should be 53 times 15 (terms of the sequence) = 795. You seem to have dropped out the 15 somewhere.
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@Aya: You did the first problem correctly .
The second one, however is wrong. By the way, are you trying to find the 8-th term of the series? Or the sum of the first 8 terms?
Assuming that you want to find the 8-th term, there's a small error in your work: you seem to forget a minus sign.

^ ok, thanks for all your help