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The 25th term of an arithmetic series is 44.5

The sum of the first 30 terms of this arithmetic series is 765

Find the 16th term of the arithmetic series.

Show your working clearly.

User Durgesh Chaudhary
by
2.9k points

1 Answer

25 votes
25 votes

Answer:


a_(16)=26.5

Explanation:

Arithmetic series:
a_n=a+(n-1)d

Sum of arithmetic series:
S_n=(n)/(2)[2a+(n-1)d]

(where a is the first term and d is the common difference)

Given:


  • a_(25)=44.5

  • S_(30)=765


\implies a_(25)=44.5


\implies a+(25-1)d=44.5


\implies a+24d=44.5


\implies S_(30)=765


\implies (30)/(2)[2a+(30-1)d]=765


\implies 30a+435d=765

Rewrite
a+24d=44.5 to make a the subject:


\implies a=44.5-24d

Substitute found expression for a into
30a+435d=765 and solve for d:


\implies 30(44.5-24d)+435d=765


\implies 1335-285d=765


\implies 285d=570


\implies d=2

Substitute found value of d into
a=44.5-24d and solve for a:


\implies a=44.5-24(2)=-3.5

Therefore:


\implies a_(16)=-3.5+(16-1)2=26.5

User Edgar Froes
by
3.0k points
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