From the question it is given that,

First term a = \[17\]

Common difference = \[14\text{ }\text{ }17\text{ }=\text{ }\text{ }3\]

Last term l = \[-40\]

\[\begin{array}{*{35}{l}}

L\text{ }=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d  \\

-40\text{ }=\text{ }17\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\left( -3 \right)  \\

-40\text{ }\text{ }17\text{ }=\text{ }-3n\text{ }+\text{ }3  \\

\text{ }57\text{ }\text{ }3\text{ }=\text{ }-3n  \\

n\text{ }=\text{ }-60/-3  \\

n\text{ }=\text{ }20  \\

\end{array}\]

Therefore, \[{{6}^{th}}\] term form the end = l – (n – 1)d

\[\begin{array}{*{35}{l}}

   =\text{ }\text{ }40\text{ }\text{ }\left( 6\text{ }\text{ }1 \right)\left( -3 \right)  \\

   =\text{ }\text{ }40\text{ }\text{ }\left( 5 \right)\left( -3 \right)  \\

   =\text{ }\text{ }40\text{ }+\text{ }15  \\

   =\text{ }\text{ }25  \\

\end{array}\]