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,

{{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.