Insert six A.M.s between 15 and −13.

Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ be the 6 A.M.s between 15 and $-13$.

Then, $15, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ and $-13$ are in A.P. whose common difference is as follows:

$d=\frac{-13-15}{6+1}$

=-4

$A_{1}=15+d=15+(-4)=11$

$A_{2}=15+2 d=15+(-8)=7$

$A_{3}=15+3 d=15+(-12)=3$

$A_{4}=15+4 d=15+(-16)=-1$

$A_{5}=15+5 d=15+(-20)=-5$

$A_{6}=15+6 d=15+(-24)=-9$

Hence, the required A.M.s are $11,7,3,-1,-5,-9$.