Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# 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

1) a4 – a1

2) [a1 + a4] / 2

3) 1

4) [a2 + a3] / 2

5) 0

Solution: (5) 0

\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}