The numbers `t (t ^(2) +1) , -1/2t ^(2)` and 6 are three consecutive terms of an A.P. If t be real, then find the the next two terms of A.P.

Answer

To solve the problem, we need to determine the values of \( t \) such that the numbers \( t(t^2 + 1) \), \( -\frac{1}{2}t^2 \), and \( 6 \) are three consecutive terms of an arithmetic progression (A.P.).

### Step-by-step Solution:

1. **Understanding the condition for A.P.**:
For three numbers \( a, b, c \) to be in A.P., the condition is:
\[ 2b = a + c \]
Here, let:
- \( a = t(t^2 + 1) \)
- \( b = -\frac{1}{2}t^2 \)
- \( c = 6 \)

2. **Setting up the equation**:
Substitute the values into the A.P. condition:
\[ 2\left(-\frac{1}{2}t^2\right) = t(t^2 + 1) + 6 \]
Simplifying the left side:
\[ -t^2 = t(t^2 + 1) + 6 \]

3. **Rearranging the equation**:
Move all terms to one side:
\[ -t^2 - t(t^2 + 1) - 6 = 0 \]
Expanding the equation:
\[ -t^2 - t^3 - t - 6 = 0 \]
Rearranging gives:
\[ t^3 + t^2 + t + 6 = 0 \]

4. **Finding the roots of the polynomial**:
We can try \( t = -2 \) as a potential root:
\[ (-2)^3 + (-2)^2 + (-2) + 6 = -8 + 4 - 2 + 6 = 0 \]
Thus, \( t = -2 \) is indeed a root.

5. **Finding the next two terms of the A.P.**:
Substitute \( t = -2 \) back into the terms:
- First term \( a = t(t^2 + 1) = -2((-2)^2 + 1) = -2(4 + 1) = -2 \times 5 = -10 \)
- Second term \( b = -\frac{1}{2}(-2)^2 = -\frac{1}{2}(4) = -2 \)
- Third term \( c = 6 \)

6. **Finding the common difference \( d \)**:
The common difference \( d \) can be calculated as:
\[ d = b - a = -2 - (-10) = 8 \]

7. **Calculating the next two terms**:
- Fourth term \( a + 3d = -10 + 3 \times 8 = -10 + 24 = 14 \)
- Fifth term \( a + 4d = -10 + 4 \times 8 = -10 + 32 = 22 \)

### Final Result:
The next two terms of the A.P. are **14** and **22**.