find the value of x such that 1+4+7+10+….+x = 715

Explanation

To solve the problem, we need to find the value of \( x \) such that the sum of the arithmetic progression (AP) \( 1 + 4 + 7 + 10 + \ldots + x = 715 \).

### Step-by-step Solution:

1. **Identify the First Term and Common Difference**:
- The first term \( a = 1 \).
- The common difference \( d = 4 - 1 = 3 \).

2. **Find the Number of Terms (n)**:
- The \( n \)-th term of an AP can be expressed as:
\[ a_n = a + (n - 1) d \]
- We know that the last term \( x \) can be expressed as:
\[ x = 1 + (n - 1) \cdot 3 \]
- Rearranging gives:
\[ x = 3n - 2 \]

3. **Use the Sum Formula for an AP**:
- The sum \( S_n \) of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \cdot (a + l) \]
- Here, \( l \) is the last term, which we have expressed as \( x \).
- Thus, we can write:
\[ S_n = \frac{n}{2} \cdot (1 + x) = 715 \]

4. **Substituting for x**:
- Substitute \( x = 3n - 2 \) into the sum formula:
\[ S_n = \frac{n}{2} \cdot (1 + (3n - 2)) = \frac{n}{2} \cdot (3n - 1) \]
- Setting this equal to 715:
\[ \frac{n}{2} \cdot (3n - 1) = 715 \]

5. **Clear the Fraction**:
- Multiply both sides by 2:
\[ n(3n - 1) = 1430 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 3n^2 - n - 1430 = 0 \]

7. **Using the Quadratic Formula**:
- The quadratic formula is given by:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 3 \), \( b = -1 \), and \( c = -1430 \).
- Calculate the discriminant:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 3 \cdot (-1430) = 1 + 17160 = 17161 \]
- Now, apply the quadratic formula:
\[ n = \frac{-(-1) \pm \sqrt{17161}}{2 \cdot 3} = \frac{1 \pm 131}{6} \]
- This gives us two possible values for \( n \):
\[ n = \frac{132}{6} = 22 \quad \text{and} \quad n = \frac{-130}{6} \quad \text{(not valid since n must be positive)} \]

8. **Finding the Last Term (x)**:
- Now that we have \( n = 22 \), substitute back to find \( x \):
\[ x = 3n - 2 = 3 \cdot 22 - 2 = 66 - 2 = 64 \]

### Final Answer:
The value of \( x \) is \( 64 \).