If 100 times the `100^(t h)` term of an AP with non zero common difference equals the 50 times its `50^(t h)` term, then the `150^(t h)` term of this AP is (1) ` 150` (2) 150 times its `50^(t h)` term (3) 150 (4) zero

Options

A. `-150`
B. 150 times its 50 th term
C. 150
D. Zero

Explanation

To solve the problem, we need to analyze the given information about the arithmetic progression (AP) and find the 150th term based on the conditions provided.

### Step-by-Step Solution

1. **Understanding the Terms of the AP**:
The nth term of an AP can be expressed as:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Expressing the Given Conditions**:
We are given that:
\[ 100 \times T_{100} = 50 \times T_{15} \]
Substituting the formula for the terms:
\[ 100 \times (a + 99d) = 50 \times (a + 14d) \]

3. **Simplifying the Equation**:
Expanding both sides:
\[ 100a + 9900d = 50a + 700d \]
Rearranging the equation to isolate terms involving \( a \) and \( d \):
\[ 100a - 50a + 9900d - 700d = 0 \]
This simplifies to:
\[ 50a + 9200d = 0 \]

4. **Finding the Relationship Between \( a \) and \( d \)**:
From the equation \( 50a + 9200d = 0 \), we can express \( a \) in terms of \( d \):
\[ a = -\frac{9200}{50}d = -184d \]

5. **Finding the 150th Term**:
Now we need to find the 150th term \( T_{150} \):
\[ T_{150} = a + (150-1)d = a + 149d \]
Substituting \( a = -184d \):
\[ T_{150} = -184d + 149d = -35d \]

6. **Conclusion**:
Since \( d \) is a non-zero common difference, \( T_{150} \) can be expressed as:
\[ T_{150} = -35d \]
However, we need to check if this term can be zero. Since \( d \) is non-zero, \( T_{150} \) will not be zero unless \( d \) is specifically chosen to make it zero, which contradicts the problem statement. Thus, we conclude that:
\[ T_{150} = 0 \text{ if } d = 0 \text{ (but } d \text{ is non-zero)} \]
Therefore, the answer is:
\[ \text{The } 150^{th} \text{ term of the AP is } 0. \]

### Final Answer:
The 150th term of this AP is (4) zero.