Question

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.\[m + n + k\]

Answer 1

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.