If x, y, z are in arithmetic progression with common difference d, x≠3d, and the determinant of the matrix [[3, 4√2, x], [4, 5√2, y], [5, k, z]] is zero, then the value of k² is:
If x, y, z are in arithmetic progression with common difference d, x≠3d, and the determinant of the matrix [[3, 4√2, x], [4, 5√2, y], [5, k, z]] is zero, then the value of k² is:
Option 1 -
6
Option 2 -
72
Option 3 -
36
Option 4 -
12
-
1 Answer
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Correct Option - 1
Detailed Solution:Given that x, y, z are in A.P., so 2y = x + z.
The determinant is:
| 3 4√2 x |
| 4 5√2 y | = 0
| 5 k z |Apply the operation R? → R? + R? - 2R?:
The first row becomes:
(3 + 5 - 24) (4√2 + k - 25√2) (x + z - 2y)
= 0 (k - 6√2) (0)
So the determinant becomes:
| 0 k-6√2 0 |
| 4 5√2 y | = 0
| 5 k z |Expanding along the first row:
-(k - 6√2)(4z - 5y) = 0.
This implies k - 6√2 = 0 or 4z - 5y = 0.
k = 6√2 or y = 4z/5.
The condition y = 4z/5 is stated as not possible.
Therefore, k = 6√2, which means k² = (6√2)² = 36 * 2 = 72.
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