How about we consider the amount of \[n\]terms of the given A.P. as \[\text{ }25.\]

We realized that,

\[{{S}_{n}}~=\text{ }n/2\text{ }\left[ 2a\text{ }+\text{ }\left( n-1 \right)d \right]\]

where\[~n\text{ }=\text{ }number\text{ }of\text{ }terms,\text{ }a\text{ }=\text{ }initial\text{ }term,\text{ }and\text{ }d\text{ }=\text{ }normal\text{ }distinction\]

So here\[,\text{ }a\text{ }=\text{ }\text{ }6\]

\[d\text{ }=\text{ }-\text{ }11/2\text{ }+\text{ }6\text{ }=\text{ }\left( -\text{ }11\text{ }+\text{ }12 \right)/2\text{ }=\text{ }1/2\]

Consequently, we have