Solve the equation (i) `1+6+11+16+……………+x=148`
(ii) `2+5+8+11+………+x=345`

Answer

To solve the given equations step by step, we will analyze each part separately.

### Part (i): Solve `1 + 6 + 11 + 16 + ... + x = 148`

1. **Identify the sequence**: The sequence is an arithmetic progression (AP) where:
- First term \( a = 1 \)
- Common difference \( d = 5 \)

2. **Find the nth term**: The nth term of an AP can be expressed as:
\[ a_n = a + (n-1)d \]
Here, the last term \( x \) can be expressed as:
\[ x = 1 + (n-1) \cdot 5 \implies x = 5n - 4 \]

3. **Find the sum of the first n terms**: The sum \( S_n \) of the first n terms of an AP is given by:
\[ S_n = \frac{n}{2} \cdot (a + l) = \frac{n}{2} \cdot (1 + x) \]
Setting this equal to 148:
\[ \frac{n}{2} \cdot (1 + x) = 148 \]

4. **Substituting for x**: Substitute \( x = 5n - 4 \) into the sum equation:
\[ \frac{n}{2} \cdot (1 + (5n - 4)) = 148 \]
Simplifying gives:
\[ \frac{n}{2} \cdot (5n - 3) = 148 \]
\[ n(5n - 3) = 296 \]
\[ 5n^2 - 3n - 296 = 0 \]

5. **Solve the quadratic equation**: Use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5, b = -3, c = -296 \)
\[ n = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 5 \cdot (-296)}}{2 \cdot 5} \]
\[ n = \frac{3 \pm \sqrt{9 + 5920}}{10} \]
\[ n = \frac{3 \pm \sqrt{5929}}{10} \]
\[ n = \frac{3 \pm 77}{10} \]
This gives \( n = 8 \) (since \( n \) must be positive).

6. **Find x**: Substitute \( n = 8 \) back to find \( x \):
\[ x = 5(8) - 4 = 40 - 4 = 36 \]

### Part (ii): Solve `2 + 5 + 8 + 11 + ... + x = 345`

1. **Identify the sequence**: The sequence is an arithmetic progression where:
- First term \( a = 2 \)
- Common difference \( d = 3 \)

2. **Find the nth term**: The nth term can be expressed as:
\[ x = 2 + (n-1) \cdot 3 \implies x = 3n - 1 \]

3. **Find the sum of the first n terms**: The sum \( S_n \) is given by:
\[ S_n = \frac{n}{2} \cdot (a + l) = \frac{n}{2} \cdot (2 + x) \]
Setting this equal to 345:
\[ \frac{n}{2} \cdot (2 + x) = 345 \]

4. **Substituting for x**: Substitute \( x = 3n - 1 \):
\[ \frac{n}{2} \cdot (2 + (3n - 1)) = 345 \]
Simplifying gives:
\[ \frac{n}{2} \cdot (3n + 1) = 345 \]
\[ n(3n + 1) = 690 \]
\[ 3n^2 + n - 690 = 0 \]

5. **Solve the quadratic equation**: Use the quadratic formula:
- Here, \( a = 3, b = 1, c = -690 \)
\[ n = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-690)}}{2 \cdot 3} \]
\[ n = \frac{-1 \pm \sqrt{1 + 8280}}{6} \]
\[ n = \frac{-1 \pm \sqrt{8281}}{6} \]
\[ n = \frac{-1 \pm 91}{6} \]
This gives \( n = 15 \) (since \( n \) must be positive).

6. **Find x**: Substitute \( n = 15 \) back to find \( x \):
\[ x = 3(15) - 1 = 45 - 1 = 44 \]

### Final Answers:
- For part (i), \( x = 36 \)
- For part (ii), \( x = 44 \)