Question:
If \(a,b,c\) are respectively the \(5^{th}, 8^{th}, 13^{th}\) terms of an arithmetic progression, then
\[
\begin{vmatrix}
a & 5 & 1\\
b & 8 & 1\\
c & 13 & 1
\end{vmatrix}
=
\]
If \(a,b,c\) are respectively the \(5^{th}, 8^{th}, 13^{th}\) terms of an arithmetic progression, then
\[
\begin{vmatrix}
a & 5 & 1\\
b & 8 & 1\\
c & 13 & 1
\end{vmatrix}
=
\]
Show Hint
If three points are collinear, then the determinant formed using their coordinates is always equal to zero.
Updated On: Jun 22, 2026
- \(0\)
- \(1\)
- \(abc\)
- \(520\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Express the terms of the arithmetic progression.
Let the first term of the arithmetic progression be \(A\) and the common difference be \(d\).
Then,
\[ a=A+4d \] \[ b=A+7d \] \[ c=A+12d \] since \(a,b,c\) are respectively the \(5^{th},8^{th},13^{th}\) terms of the A.P.
Step 2: Observe the relation among \(a,b,c\).
Compute: \[ b-a=(A+7d)-(A+4d)=3d \] and \[ c-b=(A+12d)-(A+7d)=5d \] Hence, \[ \frac{b-a}{8-5} = \frac{c-b}{13-8} = d \] Therefore, the points \[ (a,5),\;(b,8),\;(c,13) \] are collinear.
Step 3: Use determinant property.
The determinant \[ \begin{vmatrix} a & 5 & 1\\ b & 8 & 1\\ c & 13 & 1 \end{vmatrix} \] represents twice the area of the triangle formed by the points \[ (a,5),\;(b,8),\;(c,13) \] Since the three points are collinear, the area of the triangle is zero.
Therefore, \[ \begin{vmatrix} a & 5 & 1\\ b & 8 & 1\\ c & 13 & 1 \end{vmatrix} =0 \]
Step 4: Final conclusion.
Hence, \[ \boxed{0} \] which corresponds to option (1).
Let the first term of the arithmetic progression be \(A\) and the common difference be \(d\).
Then,
\[ a=A+4d \] \[ b=A+7d \] \[ c=A+12d \] since \(a,b,c\) are respectively the \(5^{th},8^{th},13^{th}\) terms of the A.P.
Step 2: Observe the relation among \(a,b,c\).
Compute: \[ b-a=(A+7d)-(A+4d)=3d \] and \[ c-b=(A+12d)-(A+7d)=5d \] Hence, \[ \frac{b-a}{8-5} = \frac{c-b}{13-8} = d \] Therefore, the points \[ (a,5),\;(b,8),\;(c,13) \] are collinear.
Step 3: Use determinant property.
The determinant \[ \begin{vmatrix} a & 5 & 1\\ b & 8 & 1\\ c & 13 & 1 \end{vmatrix} \] represents twice the area of the triangle formed by the points \[ (a,5),\;(b,8),\;(c,13) \] Since the three points are collinear, the area of the triangle is zero.
Therefore, \[ \begin{vmatrix} a & 5 & 1\\ b & 8 & 1\\ c & 13 & 1 \end{vmatrix} =0 \]
Step 4: Final conclusion.
Hence, \[ \boxed{0} \] which corresponds to option (1).
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