Official solution
Let common difference be d.
\[ \because \] a1
+ a2
+ a3
+ ... + a11 = 0
\[ \therefore \] \[{{11} \over 2}\left[ {2{a_1} + 10d} \right]\] = 0
\[ \Rightarrow \] a1
+ 5d = 0
\[ \Rightarrow \] d = \[{ - {{{a_1}} \over 5}}\] .....(1)
Now a1
+ a3
+ a5
+ ... + a23
= (a1 + a23) \[ \times \] \[{{12} \over 2}\]
= (a1 + a1 + 22d) \times 6
= \[\left[ {2{a_1} + 22\left( { - {{{a_1}} \over 5}} \right)} \right]\] \[ \times \] 6
= \[ - {{72} \over 2}{a_1}\]
\[ \therefore \] k = \[ - {{72} \over 2}\]