Question

If $ a,\,\,b,\,\,c, $ are respectively the $ p^{th},\,\,q^{th},\,\,r^{th} $ terms of an $ AP $ , then $ \left[ \begin{matrix} a & p & 1 \\ b & q & 1 \\ c & r & 1 \\ \end{matrix} \right] $ is equal to:

• A. 1
• B. -1
• C. 0
• D. pqr

Answer 1

Answer: (C)

Let the first term be $ a_{1} $ and common difference be $ d. $
According to the condition $ T_{p}=a_{1}+(p-1)d $
$ \Rightarrow $ $ a=a_{1}+(p-1)d $ ... (i)
$ T_{q}=a_{1}+(q-1)d $
$ \Rightarrow $ $ b=a_{1}+(q-1)d $ ... (ii)
and $ T_{r}=a_{1}+(r-1)d $
$ \Rightarrow $ $ c=a_{1}+(r-1)d $ ... (iii)
$ \therefore $ $ \left| \begin{matrix} a & p & 1 \\ b & q & 1 \\ c & r & 1 \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & +(p-1)d & p\,\,\,1 \\ {{a}_{1}} & +(q-1)d & p\,\,\,1 \\ {{a}_{1}} & +(r-1)d & r\,\,\,1 \\ \end{matrix} \right| $
Applying $ C_{1}\to C_{1}-(C_{2}-C_{1})d $
$ =\left| \begin{matrix} {{a}_{1}} & p & 1 \\ {{a}_{1}} & q & 1 \\ {{a}_{1}} & r & 1 \\ \end{matrix} \right| $
$ ={{a}_{1}}\left| \begin{matrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \\ \end{matrix} \right| $
$ =0[\because $two columns are identical]
Note: If any two rows or columns are identical or proportional, then the value of determinant will be zero: