## Finding Averages of the A.P.

Average of an A.P. and Corresponding terms of the A.P.
Consider the A.P., 2, 6, 10, 14, 18, 22. If you try to find the average of these six numbers you will get: Average = (2 + 6 + 10 + 14 + 18 + 22)/ 6 = 12 Notice that 12 is also the average of the first and the last terms of the A.P. In fact, it is also the average of 6 and 18 (which correspond to the second and 5th terms of the A.P.). Further, 12 is also the average of the 3rd and 4th terms of the A.P.
(Note: In this A.P. of six terms, the average was the same as the average of the 1st and 6th terms. It was also given by the average of the 2nd and the 5th terms, as well as that of the 3rd and 4th terms. )
We can call each of these pairs as “CORRESPONDING TERMS” in an A.P.
What you need to understand is that every A.P. has an average.
And for any A.P., the average of any pair of corresponding terms will also be the average of the A.P.
If you try to notice the sum of the term numbers of the pair of corresponding terms given above:
1st and 6th (so that 1 + 6 = 7)
2nd and 5th( hence, 2 + 5 = 7)
3rd and 4th (hence, 3 + 4 = 7)
Note that: In each of these cases, the sum of the term numbers for the terms in a corresponding pair is one greater than the number of terms of the A.P.
This rule will hold true for all A.P.s.
For example, if an A.P. has 23 terms then for instance, you can predict that the 7th term will have the 17th term as its corresponding term, or for that matter the 9th term will have the 15th term as its corresponding term. (Since 24 is one more than 23 and 7 + 17 = 9 + 15 = 24.)
3. Process for finding the sum of an A.P.
Once you can find a pair of corresponding terms for any A.P., you can easily find the sum of the A.P. by using the property of averages:
i.e. Sum = Number of terms ¥ Average.
In fact, this is the best process for finding the sum of an A.P. It is much more superior than the process of finding the sum of an A.P. using the expression n(2a +( n– 1) d)/2.
4. Finding the common difference of an A.P., given 2 terms of an A.P.
Suppose you were given that an A.P. had its 3rd term as 8 and its 8th term as 28. You should visualize this A.P. as – , – , 8 , – , – , –, – , 28. From the above figure, you can easily visualize that to move from the third term to the eighth term, (8 to 28) you need to add the common difference five times. The net addition being 20, the common difference should be 4.
Illustration: Find the sum of an A.P. of 17 terms, whose 3rd term is 8 and 8th term is 28.
Solution: Since we know the third term and the eighth term, we can find the common difference as 4 by the process illustrated above. The total = 17 ¥ Average of the A.P.
Our objective now shifts into the finding of the average of the A.P. In order to do so, we need to identify either the 10th term (which will be the corresponding term for the 8th term) or the 15th term (which will be the corresponding term for the 3rd term.)
Again: Since the 8th term is 28 and d = 4, the 10th term becomes 28 + 4 + 4 = 36.
Thus, the average of the A.P. = Average of 8th and 10th terms = (28 + 36)/ 2 = 32.
Hence, the required answer is sum of the A.P. = 17 ¥ 32 = 544.
The logic that has applied here is that the difference in the term numbers will give you the number of times the common difference is used to get from one to the other term.
For instance, if you know that the difference between the 7th term and 12th term of an AP is –30, you should realize that 5 times the common difference will be equal to –30. (Since 12 – 7 = 5).
Hence, d = – 6.

Understanding the logical aspects about Arithmetic Progressions is likely to help you solve questions based on APs in the context of an aptitude exam like GRE.

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