If a , b , c are respectively the 5 th , 8 th , 13 th terms of an arithmetic progression, then a 5 1 b 8 1 c…

Since, $a,b,c$ are the $5^{\mathrm{th} },8^{\mathrm{th} }$ an ${13}^{\mathrm{th} }$ term of an $\mathrm{A}.\mathrm{P}.$, respectively, therefore

$a=A+4D$

$b=A+7D$

$c=A+12D$

whew, $A$ and $D$ awe the first term and common differince of an $\mathrm{A}.\mathrm{P}.$.

Now,

$\left|\begin{array}{ccc}a& 5& 1\\ b& 8& 1\\ c& 13& 1\end{array}\right|$

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$=\left|\begin{array}{ccc}a& 5& 1\\ b-a& 3& 0\\ c-a& 8& 0\end{array}\right|$

$=8b-8a-3c+3a$

$=D\left(56-32-36+12\right)$

$=0$