Solution:
I. c, is the arithmetic mean of a and b ⇒c lies in between a and b. And it lies exactly in between the two terms. As in the number of terms between a and c should be equal to number of terms between b and c. a,c, b could be the 1st, 2nd and 3rd termsrespectively, or the 1st, 3rd and 5th, respectively, or the 2nd, 3rd, 4th respectively, or the 3rd, 4th, 5th, respectively. The terms could also be the other way around. As in, b, c, a could be the 1st 2nd and 3rd terms respectively, or the 1st, 3rd and5th respectively and so on This is a very simple but very powerful idea.
II Now, d is the arithmetic mean of b and c ⇒ d lies between b and c. Using statements I and II we can say that a.c, fc> have to be 1st. 3rd and 5th or 5th, 3rd and 1st as thereis an element between b and c also So, c is the third term, a and b are 1 st and 5th in some order.
Statement I : The average of all 5 terms put together is c. c is the middle term. So this is true.
Statement II: The average of cf and e is notgreater than average of a and b. Average of a, b is c. d and e are the 2nd and 4th terms of this sequence (in some order). So. their average should also be equal toe. So, both these are equal. So, this statement is also true.
Statement III :The average of b and c is greater than average of a and d. The average of b and c is d. The average of a and d could be greater than or less than cf. So, this need not be true.
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