Question

If we have \[{{t}_{5}},\text{ }{{t}_{10}}\text{ and }{{\text{t}}_{\text{25}}}\text{ are }{{\text{5}}^{\text{th}}}\text{, 1}{{\text{0}}^{\text{th}}}\text{ and 2}{{\text{5}}^{\text{th}}}\text{ }\] terms of AP respectively, then the value of
\[\left| \begin{matrix}
   {{t}_{5}} & {{t}_{10}} & {{t}_{25}} \\
   5 & 10 & 25 \\
   1 & 1 & 1 \\
\end{matrix} \right|\]is
$\begin{align}
  & A.\text{ -40} \\
 & \text{B}\text{. 1} \\
 & C.-1 \\
 & D.\text{ 0} \\
 & \text{E}\text{. 40} \\
\end{align}$

Answer 1

Hint: In this question we have to find out the value of determinant. In the given determinant some elements are the terms of an AP so here we use the formula of ${{n}^{th}}$term of an AP. If $a$ be the first term of an AP and $d$ be the common difference of an AP then ${{n}^{th}}$ term of an AP is given by ${{t}_{n}}=a+(n-1)d$ , so first of all we have to find the values of ${{t}_{5}},{{t}_{10}}$ and ${{t}_{25}}$ of the AP. Use the property of determinant ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$ and ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$ , and do the simplification, Once we get simplified term we have to put the value of ${{t}_{5}}=a+4d,{{t}_{10}}=a+9d$ and ${{t}_{25}}=a+24d$ After that do the algebraic operation in order to find the value of given determinant.

Complete step-by-step answer:
It is given from question that ${{t}_{5}},{{t}_{10}}$ and ${{t}_{25}}$are ${{5}^{th}},{{10}^{th}}$ and ${{25}^{th}}$ terms of the given AP
As we know that ${{n}^{th}}$ term of an AP is given by ${{t}_{n}}=a+(n-1)d$ so we can write
$\begin{align}
  & {{t}_{5}}=a+(5-1)d \\
 & {{t}_{5}}=a+4d---------(1) \\
\end{align}$
Similarly, we can write
${{t}_{10}}=a+9d-------(2)$
And
${{t}_{25}}=a+24d------(3)$
Now we have given determinant
\[\left| \begin{matrix}
   {{t}_{5}} & {{t}_{10}} & {{t}_{25}} \\
   5 & 10 & 25 \\
   1 & 1 & 1 \\
\end{matrix} \right|\]
Now operating ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$, we can write
\[\begin{align}
  & \left| \begin{matrix}
   {{t}_{5}}-{{t}_{10}} & {{t}_{10}}-{{t}_{25}} & {{t}_{25}} \\
   5-10 & 10-25 & 25 \\
   1-1 & 1-1 & 1 \\
\end{matrix} \right| \\
 & \Rightarrow \left| \begin{matrix}
   {{t}_{5}}-{{t}_{10}} & {{t}_{10}}-{{t}_{25}} & {{t}_{25}} \\
   -5 & -15 & 25 \\
   0 & 0 & 1 \\
\end{matrix} \right| \\
\end{align}\]
Now opening the determinant, we can write further
$\begin{align}
  & \left( {{t}_{5}}-{{t}_{10}} \right)\left[-15(1)-0(25) \right]-({{t}_{10}}-{{t}_{25}})[-5(1)-0(25)]+{{t}_{25}}[-5(0)-0(-15)] \\
 & \Rightarrow -15({{t}_{5}}-{{t}_{10}})+5({{t}_{10}}-{{t}_{25}}) \\
 & \Rightarrow -15{{t}_{5}}+15{{t}_{10}}+5{{t}_{10}}-5{{t}_{25}} \\
 & \Rightarrow -15{{t}_{5}}+20{{t}_{10}}-5{{t}_{25}} \\
\end{align}$
Now we have to put the values of ${{t}_{5}},{{t}_{10}}$ and ${{t}_{25}}$ from (1),(2) and (3)
So, we can write
\[\begin{align}
  & =-15(a+4d)+20(a+9d)-5(a+24d) \\
 & =-15a-60d+20a+180d-5a-120d \\
 & =-15a+15a-180d+180d \\
 & =0+0 \\
 & =0 \\
\end{align}\]
Hence the value of the determinant is zero. So, option D is correct.

Note: It should be important to note that whatever we assume in question in order to solve the question, the assumed term never comes in the final answer. All are eliminated during calculation, if they occur it means we are mistaken. In our answer the assumed term $a,d$are not in the final answer.
Also, if in a given sequence ${{t}_{1}},{{t}_{2}},{{t}_{3}},....$
${{t}_{n}}-{{t}_{n-1}}=d$(constant) then the given sequence is in AP.
Whenever we do the arithmetic operation for any column or row of a determinant with the combination of columns and rows respectively, the value of the determinant is not changed. So, we can do
$\begin{align}
  & {{C}_{1}}\to a{{C}_{1}}\pm b{{C}_{2}}\pm c{{C}_{3}} \\
 & {{R}_{1}}\to \alpha {{R}_{1}}\pm \beta {{R}_{2}}\pm \gamma {{R}_{3}} \\
\end{align}$