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Consider the set of numbers 11, 3, 6, 3, 5, 3 and x. If the mean, median and mode of this set of numbers are in an non-constant arithmetic progression. What are the number of possible values for x?
0
1
2
None of the above
It is given mean,median and mode are in A.P. Hence we have mean + mode = 2*median The mode of the set of numbers is 3. And the mean of the numbers is (x+31)/7. If x<3, the median is 3. But, as the arithmetic progression is non constant, no such values are possible. If 3 < x < 5, the median is x. In this case, (x+31)/7+3 = 2x or x = 4. This lies in between 3 and 5 and is a valid solution. If x > 5, the median is 5. So, (x+31)/7 + 3 = 10 or x = 18. As this is greater than 5, this is also a valid solution. So, total number of valid solutions is 2
By: Munesh Kumari ProfileResourcesReport error
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