To solve the problem of inserting arithmetic means between given numbers, we will follow a systematic approach for each part of the question.
### (i) Insert 5 Arithmetic Means between 8 and 26
1. **Identify the first and last terms**:
- First term (a) = 8
- Last term (l) = 26
2. **Determine the number of terms**:
- We need to insert 5 arithmetic means, so the total number of terms (n) = 5 (means) + 2 (end terms) = 7.
3. **Use the formula for the nth term of an AP**:
- The nth term of an arithmetic progression can be calculated using the formula:
\[
l = a + (n-1)d
\]
- Here, we have:
\[
26 = 8 + (7-1)d
\]
- Simplifying this gives:
\[
26 = 8 + 6d
\]
\[
6d = 26 - 8
\]
\[
6d = 18
\]
\[
d = 3
\]
4. **Calculate the arithmetic means**:
- Now we can find the arithmetic means:
- First mean (a1) = a + d = 8 + 3 = 11
- Second mean (a2) = a1 + d = 11 + 3 = 14
- Third mean (a3) = a2 + d = 14 + 3 = 17
- Fourth mean (a4) = a3 + d = 17 + 3 = 20
- Fifth mean (a5) = a4 + d = 20 + 3 = 23
5. **Final answer**:
- The 5 arithmetic means between 8 and 26 are: **11, 14, 17, 20, 23**.
### (ii) Insert 6 Arithmetic Means between 3 and 24
1. **Identify the first and last terms**:
- First term (a) = 3
- Last term (l) = 24
2. **Determine the number of terms**:
- Total number of terms (n) = 6 (means) + 2 (end terms) = 8.
3. **Use the formula for the nth term of an AP**:
- Using the formula:
\[
24 = 3 + (8-1)d
\]
- Simplifying gives:
\[
24 = 3 + 7d
\]
\[
7d = 24 - 3
\]
\[
7d = 21
\]
\[
d = 3
\]
4. **Calculate the arithmetic means**:
- Now we can find the arithmetic means:
- First mean (a1) = a + d = 3 + 3 = 6
- Second mean (a2) = a1 + d = 6 + 3 = 9
- Third mean (a3) = a2 + d = 9 + 3 = 12
- Fourth mean (a4) = a3 + d = 12 + 3 = 15
- Fifth mean (a5) = a4 + d = 15 + 3 = 18
- Sixth mean (a6) = a5 + d = 18 + 3 = 21
5. **Final answer**:
- The 6 arithmetic means between 3 and 24 are: **6, 9, 12, 15, 18, 21**.
### (iii) Insert 10 Arithmetic Means between 2 and 57
1. **Identify the first and last terms**:
- First term (a) = 2
- Last term (l) = 57
2. **Determine the number of terms**:
- Total number of terms (n) = 10 (means) + 2 (end terms) = 12.
3. **Use the formula for the nth term of an AP**:
- Using the formula:
\[
57 = 2 + (12-1)d
\]
- Simplifying gives:
\[
57 = 2 + 11d
\]
\[
11d = 57 - 2
\]
\[
11d = 55
\]
\[
d = 5
\]
4. **Calculate the arithmetic means**:
- Now we can find the arithmetic means:
- First mean (a1) = a + d = 2 + 5 = 7
- Second mean (a2) = a1 + d = 7 + 5 = 12
- Third mean (a3) = a2 + d = 12 + 5 = 17
- Fourth mean (a4) = a3 + d = 17 + 5 = 22
- Fifth mean (a5) = a4 + d = 22 + 5 = 27
- Sixth mean (a6) = a5 + d = 27 + 5 = 32
- Seventh mean (a7) = a6 + d = 32 + 5 = 37
- Eighth mean (a8) = a7 + d = 37 + 5 = 42
- Ninth mean (a9) = a8 + d = 42 + 5 = 47
- Tenth mean (a10) = a9 + d = 47 + 5 = 52
5. **Final answer**:
- The 10 arithmetic means between 2 and 57 are: **7, 12, 17, 22, 27, 32, 37, 42, 47, 52**.
