Question

If ${t_1},{t_2},{t_3},{t_4},{t_5}$ be in A.P of common difference d then the value of

{{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}$$

Answer 1

This question is based on arithmetic progression and this is also a sequence of the form $a,a + d,a + 2d,...$ where a is the first term and d is the common difference. Here in this question, we have to prove the given inequality. The above inequality is in the form of determinant by using row reduced echelon form we simplify the question.

Complete step by step solution:
The general arithmetic progression is of the form $a,a + d,a + 2d,...$ where a is first term and d is the common difference. The nth term of the arithmetic progression is defined as ${T_n} = a + (n - 1)d$
The arithmetic series is ${t_1},{t_2},{t_3},{t_4},{t_5}$ . Since it is an arithmetic series we have a common difference for the series.
The difference is represented as d. so we have
${t_2} - {t_1} = d$
Now consider the determinant

{{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|$$ To solve this, we use row transformations. On applying the row transformation $${R_3} = {R_3} - {R_2}$$ and $${R_2} = {R_2} - {R_1}$$ $$ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\\ {{t_3}{t_4} - {t_2}{t_3}}&{{t_3} - {t_2}}&{{t_2} - {t_1}} \\\ {{t_4}{t_5} - {t_3}{t_4}}&{{t_4} - {t_3}}&{{t_3} - {t_2}} \end{array}} \right|$$ On simplifying we have $$ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\\ {{t_3}({t_4} - {t_2})}&{{t_3} - {t_2}}&{{t_2} - {t_1}} \\\ {{t_4}({t_5} - {t_3})}&{{t_4} - {t_3}}&{{t_3} - {t_2}} \end{array}} \right|$$ As we know that $${t_{n + 1}} - {t_n} = d$$ and $${t_{n + 2}} - {t_n} = 2d$$ $$ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\\ {{t_3}2d}&d;&d; \\\ {{t_4}2d}&d;&d; \end{array}} \right|$$ On applying the row transformation $${R_3} = {R_3} - {R_2}$$ $$ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\\ {{t_3}2d}&d;&d; \\\ {{t_4}2d - {t_3}2d}&0&0 \end{array}} \right|$$ On further simplification we have $$ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\\ {{t_3}2d}&d;&d; \\\ {2d({t_4} - {t_3})}&0&0 \end{array}} \right|$$ As we know that $${t_{n + 1}} - {t_n} = d$$ and $${t_{n + 2}} - {t_n} = 2d$$ $$ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {{t_2}{t_3}}&{{t_2}}&{{t_1}} \\\ {{t_3}2d}&d;&d; \\\ {2{d^2}}&0&0 \end{array}} \right|$$ On simplifying the determinant, we have $$ \Rightarrow D = {t_2}{t_3}(0 - 0) - {t_2}(0 - 2{d^3}) + {t_1}(0 - 2{d^3})$$ On further simplification the above equation is written as $$ \Rightarrow D = 0 + {t_2}2{d^3} - {t_1}2{d^3}$$ Take $$2{d^3}$$ as common, the equation is written as $$ \Rightarrow D = 2{d^3}({t_2} - {t_1})$$ As we know that $${t_{n + 1}} - {t_n} = d$$ and $${t_{n + 2}} - {t_n} = 2d$$ $$ \Rightarrow D = 2{d^3}(d)$$ On multiplying we get $$ \Rightarrow D = 2{d^4}$$ hence we have proved the given inequality. **So, the correct answer is “Option A”.** **Note:** We must know about the arithmetic progression arrangement and it is based on the first term and common difference. The common difference of the arithmetic progression is defined as $${a_2} - {a_1}$$ where $${a_2}$$ represents the second term and $${a_1}$$represents the first term.