SOLUTION: Question: In an arithmetic series of 50 terms, the 17th term is 53 and the 28th term is 86. Find the sum of the series. Problem: I've gone through all the steps but I ended up w

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Question 662461: Question: In an arithmetic series of 50 terms, the 17th term is 53 and the 28th term is 86. Find the sum of the series.
Problem: I've gone through all the steps but I ended up with a different answer than what's in the back of my textbook. I'm going to type everything thing I did below, and hopefully someone can tell me where I messed up. I used \ when referring to division and / when referring to a fraction.
t17=53
t28=86
tn=a+(n-1)d
53=a+(17-1)d
53=a+16d
a=53-16d
Sn=n\2(2a+(n-1)d)
86=28/2(2a+(28-1)d)
86=14(2a+27d)
86=14(2(53-16d)+27d)
6/1/7=2(53-16d)+27d
6/1/7=106-32d+27d
-99/6/7=5d
-19/34/35=d
a=53-16(-19/34/35)
a=53-320
a=-267
Sn=n\2(2a+(n-1)d)
Sn= 50\2(2(-267)+(50-1)-19/34/35)
Sn=25(-534+(49)-19/34/34)
Sn=25(-1512/3/5)
Sn=-37815
The answer im the back of my textbook says it's suppose to be 3925
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The first part of your work is right.
term number 17 is 53 --> t%5B17%5D=53
term number n is t%5Bn%5D=a%2B%28n-1%29d
53=a%2B%2817-1%29d --> 53=a%2B16d <--> a=53-16d
You were on your way to finding a and d.
You just needed to use more data to get a and d.

Next, you misinterpreted "the 28th term is 86" (or maybe lost track of your planned steps)
You wrote S%5Bn%5D=%28n%5C2%29%282a%2B%28n-1%29d%29 , the formula for sum of the first n terms,
followed by 86=%2828%2F2%29%282a%2B%2828-1%29d%29 as if the sum of the first 28 terms was 86.
You knew that the next step would require using that formula for the sum,
but lost track, and skipped a step.

At that point, a bunch of people would write some of this:
term number 28 is 86 --> t%5B28%5D=86
term number n is t%5Bn%5D=a%2B%28n-1%29d
86=a%2B%2828-1%29d --> 86=a%2B27d
Then, they would use 86=a%2B27d, along with 53=a%2B16d to find a and d.
They would write the system system%2886=a%2B27d%2C53=a%2B16d%29
and would solve it to get highlight%28a=5%29 and highlight%28d=3%29.

RANT:
I may write all that to go with the flow, and please a teacher,
but I prefer to use definitions, logic and 5th grader reasoning (if possible),
trying to always understand the meaning of my calculations,
rather than using formulas.
I look at it the fifth grader way:
Going from the 17th to the 28th term, we are adding d from one term to the next,
and we do that 28-17=11 times,
so the difference between the 17th and the 28th terms must be 11 times d.
The difference is 86-53=33, so I divide to get d=33%2F11=highlight%283%29
Then, since you gave me a=53-16d,
I use it to get a=53-16%2A3 --> a=53-48 --> highlight%28a=5%29.

THE SUM:
S%5Bn%5D=%28n%5C2%29%282a%2B%28n-1%29d%29
S%5B50%5D=%2850%5C2%29%282%2A5%2B%2850-1%29%2A3%29 --> S%5B50%5D=%2850%5C2%29%2810%2B49%2A3%29 --> S%5B50%5D=%2850%5C2%29%2810%2B147%29 --> S%5B50%5D=%2850%5C2%29%28157%29 --> highlight%28S%5B50%5D=3925%29.