If a₁, a₂, a₃ …… are in AP, then the value of

\(\begin{vmatrix} a_{1} &a_{2} &1 \\ a_{2}&a_{3} &1 \\ a_{3}&a_{4} &1 \end{vmatrix}\)

is

\(\begin{array}{l}\begin{aligned} &\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3} \ldots \text { are in } \mathrm{AP}\\ &\therefore \mathrm{a}_{2}-\mathrm{a}_{1}=\mathrm{a}_{3}-\mathrm{a}_{2}=\mathrm{a}_{4}-\mathrm{a}_{3}=\cdots=\mathrm{d}\\ &\text { Let } \mathrm{A}=\left|\begin{array}{lll} \mathrm{a}_{1} & \mathrm{a}_{2} & 1 \\ \mathrm{a}_{2} & \mathrm{a}_{3} & 1 \\ \mathrm{a}_{3} & \mathrm{a}_{4} & 1 \end{array}\right|\\ &\text { on performing } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2} \text { we get }\\ &A=\left|\begin{array}{ccc} a_{1} & a_{2} & 1 \\ a_{2}-a_{1} & a_{3}-a_{2} & 0 \\ a_{3}-a_{2} & a_{4}-a_{3} & 0 \end{array}\right|=\left|\begin{array}{ccc} a_{1} & a_{2} & 1 \\ d & d & 0 \\ d & d & 0 \end{array}\right|=0\\ &\text { as two rows are identical } \end{aligned}\end{array} \)