Find the value of 'x' if
(i) `1+6+11+...+x=189`
`(ii) 1+1+4+7+10+...+x=287`

Answer

To solve the given problems, we will break them down into two parts as specified in the question.

### Part (i): Find 'x' such that \(1 + 6 + 11 + ... + x = 189\)

1. **Identify the sequence**: The sequence is an arithmetic progression (AP) where the first term \(a = 1\) and the common difference \(d = 6 - 1 = 5\).

2. **Find the general term**: The \(n\)-th term of an AP can be given by the formula:
\[ a_n = a + (n-1) \cdot d \]
Thus, the \(n\)-th term is:
\[ a_n = 1 + (n-1) \cdot 5 = 5n - 4 \]

3. **Sum of the first n terms**: The sum \(S_n\) of the first \(n\) terms of an AP is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
Plugging in the values:
\[ S_n = \frac{n}{2} \cdot (2 \cdot 1 + (n-1) \cdot 5) = \frac{n}{2} \cdot (2 + 5n - 5) = \frac{n}{2} \cdot (5n - 3) \]

4. **Set the sum equal to 189**:
\[ \frac{n}{2} \cdot (5n - 3) = 189 \]
Multiplying both sides by 2:
\[ n(5n - 3) = 378 \]
Rearranging gives:
\[ 5n^2 - 3n - 378 = 0 \]

5. **Solve the quadratic equation**: We can use the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b = -3\), and \(c = -378\):
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 5 \cdot (-378) = 9 + 7560 = 7569 \]
Taking the square root:
\[ \sqrt{7569} = 87 \]
Thus:
\[ n = \frac{3 \pm 87}{10} \]
The positive solution is:
\[ n = \frac{90}{10} = 9 \]

6. **Find the value of x**: Now substitute \(n\) back into the formula for the \(n\)-th term:
\[ x = a_n = 1 + (9-1) \cdot 5 = 1 + 40 = 41 \]

### Part (ii): Find 'x' such that \(1 + 1 + 4 + 7 + 10 + ... + x = 287\)

1. **Identify the sequence**: The sequence is \(1, 1, 4, 7, 10, ...\). The first term \(a = 1\) and the common difference \(d = 3\) (from 1 to 4).

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

3. **Sum of the first n terms**: The sum \(S_n\) of the first \(n\) terms is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
Plugging in the values:
\[ S_n = \frac{n}{2} \cdot (2 \cdot 1 + (n-1) \cdot 3) = \frac{n}{2} \cdot (2 + 3n - 3) = \frac{n}{2} \cdot (3n - 1) \]

4. **Set the sum equal to 287**:
\[ \frac{n}{2} \cdot (3n - 1) = 287 \]
Multiplying both sides by 2:
\[ n(3n - 1) = 574 \]
Rearranging gives:
\[ 3n^2 - n - 574 = 0 \]

5. **Solve the quadratic equation**: Using the quadratic formula:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 3 \cdot (-574) = 1 + 6888 = 6889 \]
Taking the square root:
\[ \sqrt{6889} = 83 \]
Thus:
\[ n = \frac{1 \pm 83}{6} \]
The positive solution is:
\[ n = \frac{84}{6} = 14 \]

6. **Find the value of x**: Substitute \(n\) back into the formula for the \(n\)-th term:
\[ x = a_n = 1 + (14-1) \cdot 3 = 1 + 39 = 40 \]

### Final Answers:
- For part (i), \(x = 41\)
- For part (ii), \(x = 40\)