How many terms of the A.P. 1,4,7.... are needed to give the sum 715?

Answer

To solve the problem of how many terms of the arithmetic progression (A.P.) 1, 4, 7, ... are needed to give the sum of 715, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the first term (A) and the common difference (D)**:
- The first term \( A = 1 \).
- The common difference \( D = 4 - 1 = 3 \).

2. **Let the number of terms be \( n \)**:
- We need to find \( n \) such that the sum of the first \( n \) terms equals 715.

3. **Use the formula for the sum of the first \( n \) terms of an A.P.**:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
Substituting the known values into the formula:
\[ 715 = \frac{n}{2} \times (2 \times 1 + (n - 1) \times 3) \]

4. **Simplify the equation**:
\[ 715 = \frac{n}{2} \times (2 + 3n - 3) \]
\[ 715 = \frac{n}{2} \times (3n - 1) \]

5. **Multiply both sides by 2 to eliminate the fraction**:
\[ 1430 = n(3n - 1) \]
\[ 1430 = 3n^2 - n \]

6. **Rearrange the equation into standard quadratic form**:
\[ 3n^2 - n - 1430 = 0 \]

7. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \( 3 \times -1430 = -4290 \) and add to \( -1 \).
- The factors are \( -66 \) and \( 65 \):
\[ 3n^2 - 66n + 65n - 1430 = 0 \]
\[ 3n(n - 22) + 65(n - 22) = 0 \]
\[ (n - 22)(3n + 65) = 0 \]

8. **Solve for \( n \)**:
- Setting each factor to zero gives:
- \( n - 22 = 0 \) → \( n = 22 \)
- \( 3n + 65 = 0 \) → \( n = -\frac{65}{3} \) (not acceptable since \( n \) must be positive)

9. **Conclusion**:
- The number of terms needed to give the sum of 715 is \( n = 22 \).

### Final Answer:
The number of terms needed is **22**.