To find the sum of last ten terms, we write the given AP in reverse order.
i.e, 126, 124, 122, …… 12, 10, 8
Here, first term, a = 126, common difference, d = 124 – 126 = -2
∴ S10=102[2a+(10−1)d] [∵Sn=n2[2a+(n−1)d]]
=5{2(126)+9(−2)}=5(252−18)
=5×234=1170