There are 4 numbers in a sequence in such a way that first 3 numbers have a common difference and

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16 Sep 2010 3 number in sequence

There are 4 numbers in a sequence in such a way that first 3 numbers have a common difference and
last 3 have a common ratio. If middle 2 numbers are 5 and 10 then product of first and last number is

OPtion

1) 50
2) 8
3) 24
4) 30
5) 40
6) 12
7) 28
8) 16
9) 18
10) 0

Solution

Let first number is x, then according to first given condition
5-x = 10-5, hence x = 0
Similarly let last number is y, then 10/5 = y/10, hence y = 20

Correct Option: 10 Best Solution (0)