Question 1094756: In an arithmetic sequence,the sum of the first twelve terms is 156 and 4th term is 8.Find
(a)the first term and the common difference
(b)the sum of the first sixteen terms
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Solve using logical reasoning and a basic understanding of arithmetic sequences, instead of trying to use the sometimes obscure formulas.
One of the key concepts in working with arithmetic sequences is that you can pair up the terms so that each pair has the same sum (or, if the number of terms is odd, there is one term in the middle which is half that sum).
In 12 terms of an arithmetic sequence, the pairs are
1 and 12;
2 and 11;
3 and 10;
4 and 9;
5 and 8;
6 and 7
Now let's solve your problem.
The sum of the 12 terms is 156; that means the average of all the terms is 156/12 = 13; and that means the sum of each pair is 26.
So in particular the sum of the 4th and 9th terms is 26. Since the 4th term is 8, the 9th term must be 18.
The 9th term is 5 terms after the 4th term; the 9th term is 10 more than the 4th term; that means the common difference is 10/5 = 2.
If the 4th term is 8 and the common difference is 2, then the first term (3 terms before the 4th) is 8-3(2)=2.
We have finished the first part of the problem: the first term is 2, and the common difference is 2.
For the second part of the question, we know the first term is 2 and the common difference is 2, so the n-th term is 2n. So the 16th term is 32.
Now again think of pairing up the numbers. With 16 terms, there will be 8 pairs; the sum of each pair is the same as the sum of the first and last terms, which is 34. So the sum of the first 16 terms is 8*34 = 272.
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