(i) Find the 10th term from the end of the A.P.
82, 79, 76, .... , 4
(ii) Find the 16th term from the end of the A.P.
3, 6, 9, .... 99

Answer

To solve the given problems, we will follow the steps outlined below:

### Part (i): Find the 10th term from the end of the A.P. 82, 79, 76, ..., 4

1. **Identify the first term (a) and the common difference (d)**:
- The first term \( a = 82 \)
- The common difference \( d = 79 - 82 = -3 \)

2. **Find the number of terms (n) in the A.P.**:
- The last term \( l = 4 \)
- The formula for the nth term of an A.P. is given by:
\[ l = a + (n-1)d \]
- Substituting the known values:
\[ 4 = 82 + (n-1)(-3) \]
- Rearranging gives:
\[ 4 - 82 = (n-1)(-3) \implies -78 = (n-1)(-3) \]
- Dividing both sides by -3:
\[ n - 1 = 26 \implies n = 27 \]

3. **Find the 10th term from the end**:
- The 10th term from the end is the (n - 10 + 1)th term:
\[ \text{Position} = n - 10 + 1 = 27 - 10 + 1 = 18 \]
- Now, find the 18th term:
\[ a_{18} = a + (18-1)d = 82 + (17)(-3) \]
- Calculating:
\[ a_{18} = 82 - 51 = 31 \]

### Part (ii): Find the 16th term from the end of the A.P. 3, 6, 9, ..., 99

1. **Identify the first term (a) and the common difference (d)**:
- The first term \( a = 3 \)
- The common difference \( d = 6 - 3 = 3 \)

2. **Find the number of terms (n) in the A.P.**:
- The last term \( l = 99 \)
- Using the nth term formula:
\[ l = a + (n-1)d \]
- Substituting the known values:
\[ 99 = 3 + (n-1)(3) \]
- Rearranging gives:
\[ 99 - 3 = (n-1)(3) \implies 96 = (n-1)(3) \]
- Dividing both sides by 3:
\[ n - 1 = 32 \implies n = 33 \]

3. **Find the 16th term from the end**:
- The 16th term from the end is the (n - 16 + 1)th term:
\[ \text{Position} = n - 16 + 1 = 33 - 16 + 1 = 18 \]
- Now, find the 18th term:
\[ a_{18} = a + (18-1)d = 3 + (17)(3) \]
- Calculating:
\[ a_{18} = 3 + 51 = 54 \]

### Final Answers:
- (i) The 10th term from the end of the A.P. is **31**.
- (ii) The 16th term from the end of the A.P. is **54**.