# If\[{T_m},{T_n},{T_k}\]are \[{m^{th}},{n^{th}}\] and \[{k^{th}}\]terms of an A.P. then\[\left| {\begin{array}{*{20}{c}}

{{T_m}}&m&1 \\

{{T_n}}&n&1 \\

{{T_k}}&k&1

\end{array}} \right| = \]?

A.\[1\]

B.\[ - 1\]

C.\[0\]

D.\[m + n + k\]

Last updated date: 27th Mar 2023

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Answer

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**Hint**: In this problem, arithmetic progression method is used to solve the determinant of arithmetic progression which terms \[{T_m},{T_n},{T_k}\] are \[{m^{th}},{n^{th}}\] and \[{k^{th}}\] . We use the formula for the \[{n^{th}}\] terms of arithmetic sequence is mentioned as follows, \[{T_n} = a + (n - 1)d\] .Arithmetic progression is defined as a mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as AP.

**:**

__Complete step-by-step answer__In the problem, we are given the determinant of A.P,

\[\left| {\begin{array}{*{20}{c}}

{{T_m}}&m&1 \\

{{T_n}}&n&1 \\

{{T_k}}&k&1

\end{array}} \right|\]

Comparing the formulas with the determinant of A.P, we have

For \[{m^{th}}\]term,\[{T_m} = a + (m - 1)d\]

For \[{n^{th}}\] term, \[{T_n} = a + (n - 1)d\]

For \[{k^{th}}\]term, \[{T_k} = a + (k - 1)d\]

Let \[a\] be the first term of A.P.

Let \[d\] be the common difference.

\[\left| {\begin{array}{*{20}{c}}

{{T_m}}&m&1 \\

{{T_n}}&n&1 \\

{{T_k}}&k&1

\end{array}} \right|\]

By substitute the above formula, we get

\[ \Rightarrow \left| {\begin{array}{*{20}{c}}

{a + (m - 1)d}&m&1 \\

{a + (n - 1)d}&n&1 \\

{a + (k - 1)d}&k&1

\end{array}} \right|\]

We need to perform row subtraction in further step, we get

Eliminate the last element of the first row as 0.

\[{R_1} \to {R_1} - {R_2}\]

\[ \Rightarrow \left| {\begin{array}{*{20}{c}}

{(m - n)d}&{m - n}&0 \\

{a + (n - 1)d}&n&1 \\

{a + (k - 1)d}&k&1

\end{array}} \right|\]

Eliminate the last element of the second row as 0.

\[{R_2} \to {R_2} - {R_3}\]

\[ \Rightarrow \left| {\begin{array}{*{20}{c}}

{(m - n)d}&{m - n}&0 \\

{(n - k)d}&{n - k}&0 \\

{a + (k - 1)d}&k&1

\end{array}} \right|\]

Taking common factor out from the 2nd row and 3rd column, then

\[ \Rightarrow (m - n)(n - k)\left| {\begin{array}{*{20}{c}}

d&1&0 \\

d&1&0 \\

{a + (k - 1)d}&k&1

\end{array}} \right|\]

By simplify the determinant of A.P, we get

\[ \Rightarrow (m - n)(n - k)[d(1 - 0) - 1(d - 0) + 0(dk - a - (k - 1)d]\]

By simplify in further step, we get

\[ \Rightarrow (m - n)(n - k)[d - d + 0] = 0\]

Therefore, the \[{m^{th}},{n^{th}}\]and\[{k^{th}}\]terms of an A.P. then\[\left| {\begin{array}{*{20}{c}}

{{T_m}}&m&1 \\

{{T_n}}&n&1 \\

{{T_k}}&k&1

\end{array}} \right| = 0\]

The final answer is Option(C) \[0\].

**So, the correct answer is “OPTION C”.**

**Note**: In this problem we need to find the terms of an arithmetic progression of the given determinant. Arithmetic progression is defined as a mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as AP. Here, we need to remember the formula for finding the \[{n^{th}}\] term of A.P is \[{T_n} = a + (n - 1)d\] . Where, \[a\] be the first term of A.P , \[d\] be the common difference and \[n\] be the number of terms.

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