Solve Arithmetic Series: Find Sum & Term - Check Answers!

AI Thread Summary
The discussion focuses on solving arithmetic series problems, where a user expresses confusion over their calculations for the sum of the series and the eighth term of a sequence. For the first problem, the correct sum of the series 4 + 11 + 18 + ... is clarified to be 795, with the formula used being S15 = n/2 [2a + (n - 1)d]. The second problem involves finding the eighth term of a geometric sequence, where a minor error regarding a negative sign is pointed out. Participants confirm that the user initially solved the first problem correctly but made a mistake in the second. Overall, the thread emphasizes the importance of careful calculation and understanding of series formulas.
Aya
Messages
46
Reaction score
0
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??
 
Physics news on Phys.org
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.
 
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. :smile:
-------------
@Aya: You did the first problem correctly :wink:.
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. :smile:
 
^ ok, thanks for all your help
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top