Let first term of A.P is a and common difference is d. T12​=a+11d=−13, .....(i) given ​Also given , S4​=24=24​(2a+3 d)=24⇒2a+3 d=12........(ii)​Multiply equation (i) by 2, and then subtract it from (ii), we get−19d=12−(−26)⇒d=−2then from (i) , we get a=9,Thus S10​=210​[2×9+(10−1)(−2)]=5[18−18]=0