For an increasing A.P. `a_1, a_2, a_n ifa_1=a_2+a_3+a_5=-12` and `a_1a_3a_5=80 ,` then which of the following is/are true? `a_1=-10` b. `a_2=-1` c. `a_3=-4` d. `a_5=+2`

Options

A. `a_1=-10`
B. `a_2=-1`
C. `a_3 =-4`
D. `a_5 = +2`

Explanation

To solve the problem, we need to find the values of the terms in the arithmetic progression (A.P.) given the conditions:

1. \( a_1 + a_3 + a_5 = -12 \)
2. \( a_1 a_3 a_5 = 80 \)

### Step 1: Express the terms of the A.P.
In an A.P., the \( n \)-th term can be expressed as:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- \( a_4 = a + 3d \)
- \( a_5 = a + 4d \)

### Step 2: Set up the first equation
From the first condition:
\[ a_1 + a_3 + a_5 = a + (a + 2d) + (a + 4d) = 3a + 6d = -12 \]
We can simplify this to:
\[ 3a + 6d = -12 \quad \Rightarrow \quad a + 2d = -4 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation
From the second condition:
\[ a_1 a_3 a_5 = a \cdot (a + 2d) \cdot (a + 4d) = 80 \]
Substituting \( a + 2d = -4 \) into the equation:
\[ a \cdot (-4) \cdot (a + 4d) = 80 \]
We need to express \( a + 4d \) in terms of \( a \):
From Equation 1, we have:
\[ d = \frac{-4 - a}{2} \]
Substituting \( d \) into \( a + 4d \):
\[ a + 4d = a + 4\left(\frac{-4 - a}{2}\right) = a - 8 - 2a = -8 - a \]
Thus, the equation becomes:
\[ a \cdot (-4) \cdot (-8 - a) = 80 \]
This simplifies to:
\[ 4a(8 + a) = 80 \]
Dividing both sides by 4:
\[ a(8 + a) = 20 \]
Rearranging gives us:
\[ a^2 + 8a - 20 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 8, c = -20 \):
\[ D = b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot (-20) = 64 + 80 = 144 \]
\[ a = \frac{-8 \pm \sqrt{144}}{2 \cdot 1} = \frac{-8 \pm 12}{2} \]
Calculating the two possible values:
1. \( a = \frac{4}{2} = 2 \)
2. \( a = \frac{-20}{2} = -10 \)

### Step 5: Find corresponding \( d \) values
Using \( a = -10 \):
\[ a + 2d = -4 \quad \Rightarrow \quad -10 + 2d = -4 \quad \Rightarrow \quad 2d = 6 \quad \Rightarrow \quad d = 3 \]

Using \( a = 2 \):
\[ 2 + 2d = -4 \quad \Rightarrow \quad 2d = -6 \quad \Rightarrow \quad d = -3 \]
Since the A.P. is increasing, we discard \( a = 2, d = -3 \).

### Step 6: Calculate the terms
Now, with \( a = -10 \) and \( d = 3 \):
- \( a_1 = -10 \)
- \( a_2 = -10 + 3 = -7 \)
- \( a_3 = -10 + 6 = -4 \)
- \( a_5 = -10 + 12 = 2 \)

### Conclusion
The values are:
- \( a_1 = -10 \)
- \( a_2 = -7 \)
- \( a_3 = -4 \)
- \( a_5 = 2 \)

### Verification of Options
- **Option a**: \( a_1 = -10 \) (True)
- **Option b**: \( a_2 = -1 \) (False)
- **Option c**: \( a_3 = -4 \) (True)
- **Option d**: \( a_5 = 2 \) (True)

### Final Answer
The true statements are:
- a. \( a_1 = -10 \)
- c. \( a_3 = -4 \)
- d. \( a_5 = 2 \)