After 10 years, there will have been 120 deposits. The last one earns 12%/12 = 1% interest, so is mutipied by 1.01. The one before is multiplied by 1.01². Overall, you have the sum of a geometric series of 120 terms with first term 10.10 and common ratio 1.01. That sum is given by the general formula Sn = a1·(r^n -1)/(r -1) S120 = 10.10(1.01^120 -1)/(1.01 -1) S120 = 1010·2.30038689 ≈ 2323
At the end of the 10th year (before the first deposit of the 11th year), the account balance will be $2,323
