Quote:
An arithmetic progression of upward positive integers has a difference of 2 and a mean of 20. How many such A.P.s are there whose sum<200?
A. 4
B. 5
C. 6
D. 7
E. 8
The right answer here is
E. This question may appear kind of difficult at first, but is actually very straightforward once you get the logic. Based on the info we know that the sum:
1. Has at least two numbers
2. The mean is 20
3. The difference is 2
What is the smallest sequence we can make with an average of 20 and difference of 2? Well it has to be 19 and 21 of course! Based on this, we can also say that a sequence that goes 17, 19, 21, 23 must also fit the criteria. So really, any sequence of even numbers would work, so long as there are fewer than 10 numbers in the sequence (since the sum of 10 numbers with mean 20 must be 200).
What about sequences with an odd amount of numbers in them? Well that's easier still, because the only criteria is that the middle number should be 20. 18, 20, 22 works just fine, as does 16, 18, 20, 22, 24. So really, any sequence of numbers that has fewer than ten total terms in it should work.
This is all the numbers from 2-9, so there are 8 possible such combinations.
- Matoo
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