Courses
Courses for Kids
Free study material
Free LIVE classes
More

# 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
Total views: 207.3k
Views today: 2.84k
Verified
207.3k+ views
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.

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.