Answer: First instalment ₹ 150, last instalment ₹ 40
Total amount repaid = 1000 + 140 = ₹ 1140. The instalments form an A.P. with n = 12, d = –10, \(S_{12} = 1140\).
\(S_n = \dfrac{n}{2}[2a + (n-1)d]\): \(1140 = 6[2a - 110]\), so \(2a - 110 = 190\), \(a = 150\).
Last instalment \(t_{12} = a + 11d = 150 - 110 = 40\).