(i) Find the number of terms in the A.P.
8, 12, 16, ........124
(ii) Find the number of terms in the A.P.
75, 70, 65, .....15

Answer

To solve the given problems step by step, we will use the formula for the nth term of an arithmetic progression (A.P.), which is given by:

\[ a_n = a + (n-1)d \]

where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms,
- \( a_n \) is the nth term.

### Part (i): Find the number of terms in the A.P. 8, 12, 16, ..., 124

**Step 1:** Identify the first term \( a \) and the common difference \( d \).

- First term \( a = 8 \)
- Second term = 12, so the common difference \( d = 12 - 8 = 4 \)

**Step 2:** Identify the last term \( a_n \).

- Last term \( a_n = 124 \)

**Step 3:** Use the formula for the nth term of an A.P. to find \( n \).

\[ a_n = a + (n-1)d \]
Substituting the known values:

\[ 124 = 8 + (n-1) \cdot 4 \]

**Step 4:** Simplify the equation.

\[ 124 = 8 + 4(n - 1) \]
\[ 124 = 8 + 4n - 4 \]
\[ 124 = 4 + 4n \]
\[ 124 - 4 = 4n \]
\[ 120 = 4n \]

**Step 5:** Solve for \( n \).

\[ n = \frac{120}{4} = 30 \]

Thus, the number of terms in the A.P. is **30**.

### Part (ii): Find the number of terms in the A.P. 75, 70, 65, ..., 15

**Step 1:** Identify the first term \( a \) and the common difference \( d \).

- First term \( a = 75 \)
- Second term = 70, so the common difference \( d = 70 - 75 = -5 \)

**Step 2:** Identify the last term \( a_n \).

- Last term \( a_n = 15 \)

**Step 3:** Use the formula for the nth term of an A.P. to find \( n \).

\[ a_n = a + (n-1)d \]
Substituting the known values:

\[ 15 = 75 + (n-1)(-5) \]

**Step 4:** Simplify the equation.

\[ 15 = 75 - 5(n - 1) \]
\[ 15 = 75 - 5n + 5 \]
\[ 15 = 80 - 5n \]
\[ 15 - 80 = -5n \]
\[ -65 = -5n \]

**Step 5:** Solve for \( n \).

\[ n = \frac{65}{5} = 13 \]

Thus, the number of terms in the A.P. is **13**.

### Summary of Solutions:
(i) The number of terms in the A.P. 8, 12, 16, ..., 124 is **30**.
(ii) The number of terms in the A.P. 75, 70, 65, ..., 15 is **13**.