There are 4 positive numbers k, k^2, k^3, k^4. If we subtract 12 by first number and 36 by last numb

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21 Dec 2011 AP and GP

There are 4 positive numbers k, k^2, k^3, k^4. If we subtract 12 by first number and 36 by last number, we will get an AP of 4 numbers. Calculate k?

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10) none of these

Solution

According to the given condition k-12, k^2, k^3, k^4-36 are in AP.
so k^2-(k-12) = k^3-k^2
so k^3-2k^2+k-12 = 0 (i)
equation (i) has 3 as its real root
on taking the value of k, given 4 numbers will be -9,9,27,45

Correct Option: 3 Best Solution (0)