Insert 3 arithmetic means between:
(i) 3 and 15 (ii) 5 and 21.

Answer

To insert 3 arithmetic means between the given pairs of numbers, we will follow these steps for each part of the question.

### Part (i): Insert 3 Arithmetic Means between 3 and 15

1. **Identify the first and last terms**:
- First term (a1) = 3
- Last term (a5) = 15

2. **Determine the total number of terms**:
- We have 3 arithmetic means, which means there are 5 terms in total (3 means + 2 given terms).

3. **Use the formula for the nth term of an arithmetic progression**:
\[ a_n = a + (n-1)d \]
For our case, we can express the 5th term as:
\[ a_5 = a_1 + (5-1)d \]
Substituting the known values:
\[ 15 = 3 + 4d \]

4. **Solve for d**:
\[ 15 - 3 = 4d \implies 12 = 4d \implies d = \frac{12}{4} = 3 \]

5. **Calculate the arithmetic means**:
- A1 (first mean) = a1 + d = 3 + 3 = 6
- A2 (second mean) = a1 + 2d = 3 + 2(3) = 3 + 6 = 9
- A3 (third mean) = a1 + 3d = 3 + 3(3) = 3 + 9 = 12

6. **Final sequence**:
- The sequence is: 3, 6, 9, 12, 15

### Part (ii): Insert 3 Arithmetic Means between 5 and 21

1. **Identify the first and last terms**:
- First term (a1) = 5
- Last term (a5) = 21

2. **Determine the total number of terms**:
- Again, we have 3 arithmetic means, leading to 5 terms in total.

3. **Use the formula for the nth term of an arithmetic progression**:
\[ a_n = a + (n-1)d \]
For our case, we express the 5th term:
\[ a_5 = a_1 + (5-1)d \]
Substituting the known values:
\[ 21 = 5 + 4d \]

4. **Solve for d**:
\[ 21 - 5 = 4d \implies 16 = 4d \implies d = \frac{16}{4} = 4 \]

5. **Calculate the arithmetic means**:
- A1 (first mean) = a1 + d = 5 + 4 = 9
- A2 (second mean) = a1 + 2d = 5 + 2(4) = 5 + 8 = 13
- A3 (third mean) = a1 + 3d = 5 + 3(4) = 5 + 12 = 17

6. **Final sequence**:
- The sequence is: 5, 9, 13, 17, 21

### Summary of Results:
- For the first part, the arithmetic means are: **6, 9, 12**
- For the second part, the arithmetic means are: **9, 13, 17**