Question

If $T_p , T_q,T_r$ are $p$th, $q$th and $r$th terms of an A.P then $\begin{vmatrix}T_{p}&T_{q}&T_{r}\\ p&q&r\\ 1&1&1\end{vmatrix} $ is equal to

• A. 1
• B. -1
• C. 0
• D. p + q + r.

Answer 1

Answer: (C)

$T_{p }= a+ \left(p-1\right) d; T_{q} = a + \left(q-1\right) d; $
$T_{r } =a + \left(r-1\right)d$
$\therefore $ the given determinant
$\begin{vmatrix}a+\left(p-1\right)d&a+\left(q-1\right)d&a+\left(r-1\right)d\\ p&q&r\\ 1&1&1\end{vmatrix}$
Operate $R_1 - (R_2 - R_3)d$,
we get the determinant = $\begin{vmatrix}a&a&a\\ p&q&r\\ 1&1&1\end{vmatrix} = 0$