The determinant D = \(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{

1.	The determinant D = \(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{vmatrix}\)= 0, if[(a, b, a+b) (b, c, b+c) (a+b, b+c, 0)](a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) α is a root of ax2 + 2bx + c = 0
Answer» Correct option is : (b) a, b, c are in G.P. Applying R₃ → R₃ – (R₁ + R₂ ), we get \(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{vmatrix}\)=0 ∴ a[-c(a + 2b + c) – 0] – b[-b(a + 2b + c) – 0] + (a + b) (0 – 0) = 0 ∴ (-ac + b²) (a + 2b + c) = 0 ∴ -ac + b² = 0 or a + 2b + c = 0 ∴ b² = ac ∴ a, b, c are in G.P.

The determinant D = \(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{vmatrix}\)= 0, if[(a, b, a+b) (b, c, b+c) (a+b, b+c, 0)](a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) α is a root of ax2 + 2bx + c = 0

Answer»

Correct option is : (b) a, b, c are in G.P.

Applying R₃ → R₃ – (R₁ + R₂ ), we get

\(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{vmatrix}\)=0

∴ a[-c(a + 2b + c) – 0] – b[-b(a + 2b + c) – 0] + (a + b) (0 – 0) = 0

∴ (-ac + b²) (a + 2b + c) = 0

∴ -ac + b² = 0 or a + 2b + c = 0

∴ b² = ac

∴ a, b, c are in G.P.