SOLUTION: The first term of an arithmetic progression is 6 and the fifth term is 18,find the number of term in the series having a sum of 162

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Question 1147514: The first term of an arithmetic progression is 6 and the fifth term is 18,find the number of term in the series having a sum of 162
Answer by greenestamps(13200) About Me  (Show Source):
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Given: 1st term 6; 5th term 18

The difference is 12; the 5th term is 4 terms after the first. So the common difference is 12/4 = 3

With a common difference of 3 and a first term of 6, the formula for the n-th term of the sequence is 3n+3.

The sum of n terms of an arithmetic progression is

(number of terms) times (average of all the terms)

which in an arithmetic progression is

(number of terms) times (average of first and last terms)

We have expressions for all those numbers:
number of terms: n
first term: 6
last (n-th) term: 3n+3

We want to find the number n that makes the sum 162:

n%28%286%2B%283n%2B3%29%29%2F2%29+=+162
n%28%283n%2B9%29%2F2%29+=+162
n%283n%2B9%29+=+324
n%28n%2B3%29+=+108

Formal algebra or trial and error shows n=9 (the other algebraic solution n=-12 doesn't make sense in the context of the problem).

ANSWER: The number of terms to have a sum of 162 is 9.

CHECK:
1st term: 6
9th term: 6+8*3 = 6+24 = 30
Sum of 9 terms: 9((6+30)/2) = 9(36/2) = 9(18) = 162