SOLUTION: in an arithmetic series, the second to last term is three times the fifth term, the sum of the first eight terms equals 136, and the fourteenth term equals the twentieth minus the

Algebra ->  Sequences-and-series -> SOLUTION: in an arithmetic series, the second to last term is three times the fifth term, the sum of the first eight terms equals 136, and the fourteenth term equals the twentieth minus the       Log On


   



Question 932868: in an arithmetic series, the second to last term is three times the fifth term, the sum of the first eight terms equals 136, and the fourteenth term equals the twentieth minus the second term. Determine how many terms are in the arithmetic series.
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
in an arithmetic series, the second to last term is three times the fifth term, the sum of the first eight terms equals 136, and the fourteenth term equals the twentieth minus the second term. Determine how many terms are in the arithmetic series.
Use formulas

a%5Bn%5D=a%5B1%5D%2B%28n-1%29d and S%5Bn%5D=expr%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%5E%22%22%29, 

>>the second to last term is three times the fifth term<<

[That's a little bit ambiguous, because "second to last" might be
interpreted as "second FROM the last"; however I will interpret
it as "NEXT TO the last"] 

a%5Bn-1%5D=3a%5B5%5D

a%5B1%5D%2B%28%28n-1%29%5E%22%22-1%29d%22%22=%22%223%28a%5B1%5D%2B%285-1%29d%5E%22%22%29

a%5B1%5D%2B%28n-2%29d%22%22=%22%223%28a%5B1%5D%2B4d%29

a%5B1%5D%2Bnd-2d%22%22=%22%223a%5B1%5D%2B12d%29

nd%22%22=%22%222a%5B1%5D%2B14d

>>the sum of the first eight terms equals 136<<
S%5B8%5D=expr%288%2F2%29%282a%5B1%5D%2B%28n-1%29d%5E%22%22%29
136=4%282a%5B1%5D%2B%288-1%29d%5E%22%22%29
136=4%282a%5B1%5D%2B7d%29
136=4%282a%5B1%5D%2B7d%29
136=8a%5B1%5D%2B28d
34=2a%5B1%5D%2B7d

>>and the fourteenth term equals the twentieth minus the second term.<<
a%5B14%5D=a%5B20%5D-a%5B2%5D

a%5B1%5D%2B%2814-1%29d%22%22=%22%22%28a%5B1%5D%2B%2820-1%29%5E%22%22d%29-%28a%5B1%5D%2B%282-1%29%5E%22%22d%29

a%5B1%5D%2B13d%22%22=%22%22%28a%5B1%5D%2B19d%29-%28a%5B1%5D%2B1d%29

a%5B1%5D%2B13d%22%22=%22%22a%5B1%5D%2B19d-a%5B1%5D-1d

a%5B1%5D%2B13d%22%22=%22%2218d

a%5B1%5D%22%22=%22%225d

So we have the system of equations:

system%28nd=2a%5B1%5D%2B14d%2C34=2a%5B1%5D%2B7d%2Ca%5B1%5D=5d%29 

Solve that by substitution and get:

a%5B1%5D+=+10, d+=+2,  n+=+24

There are 24 terms.

Edwin