SOLUTION: In an arithmetic series, t10=21 and t19=48 Find a, d, and S23 Thank You

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Question 928076: In an arithmetic series, t10=21 and t19=48
Find a, d, and S23

Thank You
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I assume a= first term, t%5B10%5D= tenth term, t%5B19%5D= term number 19 , and S%5B23%5D= sum of first 23 terms.
It is commonly understood that d= common difference.
t%5Bn%5D=a%2B%28n-1%29d , so
system%2821=a%2B9d%2C48=a%2B18d%29 --->48-21=18d-9d--->27=9d--->27%2F3=d--->highlight%28d=3%29
Then, system%2821=a%2B9d%2Cd=3%29 --->21=a%2B9%2A3--->21=a%2B27--->21-27=a--->highlight%28a=-6%29 .
There are different formulas to calculate sums of arithmetic sequences, but I prefer to remember that adding up the first n terms twice, grouping them in pairs "head-to-tail", t%5B0%5D%2Bt%5Bn%5D , t%5B1%5D%2Bt%5Bn-1%5D , t%5B2%5D%2Bt%5Bn-2%5D , and so on, you end up with a sum of n pairs that all add up to t%5B0%5D%2Bt%5Bn%5D, so
2S%5Bn%5D=n%28t%5B0%5D%2Bt%5Bn%5D%29-->S%5Bn%5D=%28n%28t%5B0%5D%2Bt%5Bn%5D%29%29%2F2 .
So, t%5B23%5D=-6%2B22%2A3=-6%2B66=60 and .