[tex]a_{12}=-13[/tex]
[tex]a_1+a_2+a_3+a_4=24[/tex]
Since the progression is arithmetic, we have
[tex]a_2=a_1+d[/tex]
[tex]a_3=a_2+d=a_1+2d[/tex]
[tex]a_4=a_3+d=a_1+3d[/tex]
So we have
[tex]a_1+(a_1+d)+(a_1+2d)+(a_1+3d)=4a_1+6d=24\implies2a_1+3d=12[/tex]
Continuing the pattern above, we also find that
[tex]a_{12}=a_{11}+d=\cdots=a_1+11d\implies a_1+11d=-13[/tex]
Writing [tex]a_1=-11d-13[/tex] and substituting into the first equation, we have
[tex]2(-11d-13)+3d=12\implies-19d-26=12\implies19d=-38\implies d=-2[/tex]
[tex]\implies a_1-22=-13\implies a_1=9[/tex]
Now the sum of the first ten terms is
[tex]a_1+a_2+\cdots+a_{10}=10a_1+(0+1+2+\cdots+9)d[/tex]
[tex]=10\cdot9+\dfrac{9\cdot10}2\cdot(-2)=0[/tex]
