Problem:

A certain sum of money is invested at the rate of 10% per annum compound interest, the interest compounded annually. If the difference between the interests of the third year is $1105, find the sum invested.

Solution:

Step 1: Understand the Problem
We are given that a certain sum of money is invested at a compound interest rate of 10% per annum, compounded annually. We need to find the initial sum of money invested when the difference between the interests of the third year is $1105.

Step 2: Determine the Formula
The formula to calculate compound interest is given by:

A = P(1 + r/n)^(nt)

Where:
A = the final amount or maturity value
P = the principal or initial amount
r = the annual interest rate (as a decimal)
n = the number of times that interest is compounded per year
t = the number of years

Step 3: Break Down the Given Information
From the problem, we know the following details:
- The interest rate is 10% per annum.
- The interest is compounded annually.
- The difference between the interests of the third year is $1105.

Step 4: Formulate the Equation
Let's assume the principal or initial amount invested is P. Using the compound interest formula, the amount after three years would be:

A = P(1 + 0.10/1)^(1*3) = P(1.10)^3

The interest earned after three years would be:

Interest = A - P

Since we are given that the difference between the interests of the third year is $1105, we can set up the following equation:

(P(1.10)^3 - P) - (P(1.10)^2 - P) = $1105

Step 5: Solve the Equation
Simplifying the equation, we get:

(P(1.10)^3 - P) - (P(1.10)^2 - P) = $1105
(1.331 - 1)P - (1.21 - 1)P = $1105
0.331P - 0.21P = $1105
0.121P = $1105
P = $1105 / 0.121
P ≈ $9125

Therefore, the sum invested is approximately $9125.