Question

If \[{t_1},{t_2},{t_3},{t_4},{t_5}\] be in a A.P. of common difference \[d\] then the value of \[D = \left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {{t_3}{t_4}}&{{t_3}}&{{t_2}} \\
  {{t_4}{t_5}}&{{t_4}}&{{t_3}}
\end{array}} \right|\] is \[2{d^4}\].
A. True
B. False

Answer 1

Hint: In this question we are given a statement we have to check if it is true or false. Solve the determinant using row transformation and to find its value. The common difference of the series in A.P. is equal to the difference between two successive terms. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given that the terms \[{t_1},{t_2},{t_3},{t_4},{t_5}\] are in A.P. Since the common difference of the series in A.P. is equal to the difference between two successive terms, we have
\[ \Rightarrow d = \left( {{t_5} - {t_4}} \right) = \left( {{t_4} - {t_3}} \right) = \left( {{t_3} - {t_2}} \right) = \left( {{t_2} - {t_1}} \right)\]
Also, we have \[2d = {t_5} - {t_3} = {t_4} - {t_2}\]
Now, we have to find the value of \[D = \left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {{t_3}{t_4}}&{{t_3}}&{{t_2}} \\
  {{t_4}{t_5}}&{{t_4}}&{{t_3}}
\end{array}} \right|\]
Apply Row transformation \[{R_3} \to {R_3} - {R_2}\], we have
\[
   \Rightarrow D = \left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {{t_3}{t_4}}&{{t_3}}&{{t_2}} \\
  {{t_4}\left( {{t_5} - {t_3}} \right)}&{{t_4} - {t_3}}&{{t_3} - {t_2}}
\end{array}} \right| \\
   \Rightarrow D = \left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {{t_3}{t_4}}&{{t_3}}&{{t_2}} \\
  {{t_4} \times 2d}&d&d
\end{array}} \right| \\
\]
Applying Row transformation \[{R_2} \to {R_2} - {R_1}\]
\[
   \Rightarrow D = \left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {{t_3}\left( {{t_4} - {t_2}} \right)}&{{t_3} - {t_2}}&{{t_2} - {t_1}} \\
  {{t_4} \times 2d}&d&d
\end{array}} \right| \\
   \Rightarrow D = \left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {{t_3} \times 2d}&d&d \\
  {{t_4} \times 2d}&d&d
\end{array}} \right| \\
\]
Taking as \[d\] common in 1st and 2nd row, we have
\[ \Rightarrow D = {d^2}\left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {2{t_3}}&1&1 \\
  {2{t_4}}&1&1
\end{array}} \right|\]
Applying the row transformation \[{R_3} \to {R_3} - {R_2}\]
\[
   \Rightarrow D = {d^2}\left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {2{t_3}}&1&1 \\
  {2\left( {{t_4} - {t_3}} \right)}&{1 - 1}&{1 - 1}
\end{array}} \right| \\
   \Rightarrow D = {d^2}\left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {2{t_3}}&1&1 \\
  {2d}&0&0
\end{array}} \right| \\
\]
Taking \[d\] as common in 3rd row, we have
\[ \Rightarrow D = {d^3}\left| {\begin{array}{*{20}{c}}
  {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\
  {2{t_3}}&1&1 \\
  2&0&0
\end{array}} \right|\]
Expanding the determinant with 3rd row, we get
\[
   \Rightarrow D = {d^3}\left[ {2\left( {{t_2} - {t_1}} \right) - 0 + 0} \right] \\
   \Rightarrow D = {d^3}\left( {2d} \right) \\
  \therefore D = 2{d^4} \\
\]
Therefore, the given statement is true.

So, the correct answer is “Option A”.

Note: An arithmetic progression is a sequence of numbers such that the difference of any two successive members are a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1. Students must know the formula for finding the modulus of a matrix.