Question

Three numbers \( x, y, \) and \( z \) are in arithmetic progression (A.P.). If \( x+y+z = -3 \) and \( xyz = 8 \), then \( x^2+y^2+z^2 \) is equal to:

• A. \( 9 \)
• B. \( 10 \)
• C. \( 21 \)
• D. \( 20 \)
• E. \( 1 \)

Hint:

Always assume three A.P. terms as \( a-d, a, a+d \). If there are four, use \( a-3d, a-d, a+d, a+3d \). This symmetry simplifies the algebra significantly.

Answer 1

The Correct Option is C

Concept: For three numbers in A.P., it is most efficient to assume them as \( a-d, a, \) and \( a+d \). This ensures that when they are summed, the common difference \( d \) cancels out, allowing for the immediate calculation of the middle term \( a \).

Step 1: Finding the values of the three numbers.
Sum of terms: \( (a-d) + a + (a+d) = -3 \implies 3a = -3 \implies a = -1 \). So the numbers are \( -1-d, -1, \) and \( -1+d \). Product of terms: \( (-1-d)(-1)(-1+d) = 8 \implies (-1)[(-1)^2 - d^2] = 8 \). \[ -(1 - d^2) = 8 \implies d^2 - 1 = 8 \implies d^2 = 9 \implies d = \pm 3 \] If \( d = 3 \), the numbers are \( -4, -1, 2 \). (The same set results if \( d = -3 \)).

Step 2: Calculating the sum of squares.
Using the identified numbers \( x=-4, y=-1, z=2 \): \[ x^2 + y^2 + z^2 = (-4)^2 + (-1)^2 + (2)^2 \] \[ = 16 + 1 + 4 = 21 \]