Step 1: Define the terms of the arithmetic progression. Let the first term be $a$ and the common difference be $d$. The first three terms of the arithmetic progression are: $t_1 = a$ $t_2 = a + d$ $t_3 = a + 2d$ Step 2: Use the information about the average of the first three terms to form an equation. The average of the first three terms is 56. $$ \frac{t_1 + t_2 + t_3}{3} = 56 $$ Substitute the expressions for $t_1, t_2, t_3$: $$ \frac{a + (a+d) + (a+2d)}{3} = 56 $$ $$ \frac{3a + 3d}{3} = 56 $$ Divide by 3: $$ a + d = 56 \quad (*1) $$ Step 3: Use the information about the sum of the first two terms to form another equation. The sum of the first two terms is 96. $$ t_1 + t_2 = 96 $$ Substitute the expressions for $t_1, t_2$: $$ a + (a+d) = 96 $$ $$ 2a + d = 96 \quad (*2) $$ Step 4: Solve the system of two linear equations. We have the system: 1) $a + d = 56$ 2) $2a + d = 96$ Subtract equation (1) from equation (2): $$ (2a + d) - (a + d) = 96 - 56 $$ $$ 2a + d - a - d = 40 $$ $$ a = 40 $$ Step 5: Substitute the value of $a$ into equation (*1) to find $d$. $$ a + d = 56 $$ $$ 40 + d = 56 $$ $$ d = 56 - 40 $$ $$ d = 16 $$ The first term is $a=40$ and the common difference is $d=16$. The first term is $\boxed{\text{40}}$ and the common difference is $\boxed{\text{16}}$.