To solve the problem, we will analyze each statement one by one, determining whether they are true or false.
### Statement 1:
If \( a, b, c \) are in A.P., and \( b, c, a \) are in G.P., then \( c, a, b \) are in H.P.
1. **Understanding A.P.**:
- If \( a, b, c \) are in A.P., then:
\[
2b = a + c \quad \text{(1)}
\]
2. **Understanding G.P.**:
- If \( b, c, a \) are in G.P., then:
\[
c^2 = ab \quad \text{(2)}
\]
3. **Substituting from (1) into (2)**:
- From (1), we can express \( c \) as:
\[
c = 2b - a
\]
- Substitute \( c \) into (2):
\[
(2b - a)^2 = ab
\]
4. **Expanding the equation**:
\[
4b^2 - 4ab + a^2 = ab
\]
- Rearranging gives:
\[
4b^2 - 5ab + a^2 = 0 \quad \text{(3)}
\]
5. **Finding the roots of (3)**:
- The quadratic equation can be solved using the quadratic formula:
\[
b = \frac{5a \pm \sqrt{(5a)^2 - 4 \cdot 4 \cdot a^2}}{2 \cdot 4}
\]
- This simplifies to:
\[
b = \frac{5a \pm \sqrt{25a^2 - 16a^2}}{8} = \frac{5a \pm 3a}{8}
\]
- Thus, we have two cases:
1. \( b = a \) (which implies \( a = b = c \))
2. \( b = \frac{1}{2}a \) (which implies \( a = 4b \) and \( c = 2b \))
6. **Checking H.P. condition**:
- If \( a = b = c \), then \( c, a, b \) are in H.P.
- If \( a = 4b \), \( c = 2b \), and \( b = b \), we check:
\[
\frac{1}{c} + \frac{1}{a} + \frac{1}{b} = \frac{1}{2b} + \frac{1}{4b} + \frac{1}{b} \neq \text{constant}
\]
- Thus, \( c, a, b \) are not in H.P. in this case.
**Conclusion for Statement 1**: True.
### Statement 2:
If \( a_1, a_2, a_3, \ldots, a_{100} \) are in A.P. and \( a_3 + a_{98} = 50 \), then \( a_1 + a_2 + a_3 + \ldots + a_{100} = 2500 \).
1. **Using the formula for A.P.**:
- The \( n \)-th term of an A.P. is given by:
\[
a_n = a + (n-1)d
\]
- Thus:
\[
a_3 = a + 2d \quad \text{and} \quad a_{98} = a + 97d
\]
- Therefore:
\[
a_3 + a_{98} = (a + 2d) + (a + 97d) = 2a + 99d = 50 \quad \text{(4)}
\]
2. **Finding the sum of the series**:
- The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[
S_n = \frac{n}{2} \times [2a + (n-1)d]
\]
- For \( n = 100 \):
\[
S_{100} = \frac{100}{2} \times [2a + 99d] = 50 \times (2a + 99d)
\]
3. **Substituting from (4)**:
- We know \( 2a + 99d = (2a + 99d) - d = 50 - d \).
- Therefore:
\[
S_{100} = 50 \times 50 = 2500
\]
**Conclusion for Statement 2**: True.
### Statement 3:
If \( a_1, a_2, \ldots, a_n \neq 0 \), then:
\[
\frac{a_1 + a_2 + \ldots + a_n}{n} \geq (a_1 a_2 \ldots a_n)^{1/n}
\]
1. **Understanding the inequality**:
- This is the AM-GM inequality, which states that the arithmetic mean is greater than or equal to the geometric mean.
- However, the condition is that \( a_i \) must be positive for AM-GM to hold.
- The statement only states \( a_i \neq 0 \), which does not guarantee positivity.
**Conclusion for Statement 3**: False.
### Final Conclusion:
- Statement 1: True
- Statement 2: True
- Statement 3: False
The most appropriate option is the second one, which states that the first two statements are true while the third is false.
