If the sum of three numbers in a GP. is 26 and the sum of products taken two at a time is 156, then the numbers are

Options

A. 1,5,25
B. 2,6,18
C. 1,4,16
D. 2,8,16

Answer

To solve the problem, we need to find three numbers in a geometric progression (GP) given that their sum is 26 and the sum of the products taken two at a time is 156.

Let the three numbers in GP be represented as:
- First number: \( \frac{a}{r} \)
- Second number: \( a \)
- Third number: \( ar \)

### Step 1: Set up the equations
From the problem, we have two equations based on the information provided:

1. The sum of the three numbers:
\[ \frac{a}{r} + a + ar = 26 \]

2. The sum of the products taken two at a time:
\[ \left(\frac{a}{r} \cdot a\right) + \left(\frac{a}{r} \cdot ar\right) + \left(a \cdot ar\right) = 156 \]

### Step 2: Simplify the equations
Let's simplify the first equation:
\[ \frac{a}{r} + a + ar = 26 \]
Multiplying through by \( r \) to eliminate the fraction:
\[ a + ar + a r^2 = 26r \]
This can be rearranged to:
\[ a(1 + r + r^2) = 26r \quad \text{(1)} \]

Now, simplify the second equation:
\[ \frac{a^2}{r} + a^2 + a^2 r = 156 \]
Factoring out \( a^2 \):
\[ a^2 \left(\frac{1}{r} + 1 + r\right) = 156 \]
Multiplying through by \( r \):
\[ a^2(1 + r + r^2) = 156r \quad \text{(2)} \]

### Step 3: Divide the equations
Now we can divide equation (2) by equation (1):
\[ \frac{a^2(1 + r + r^2)}{a(1 + r + r^2)} = \frac{156r}{26r} \]
This simplifies to:
\[ \frac{a}{1} = 6 \]
Thus, we find:
\[ a = 6 \]

### Step 4: Substitute back to find \( r \)
Now substitute \( a = 6 \) back into equation (1):
\[ 6(1 + r + r^2) = 26r \]
Dividing both sides by 6:
\[ 1 + r + r^2 = \frac{26r}{6} = \frac{13r}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 3 + 3r + 3r^2 = 13r \]
Rearranging gives:
\[ 3r^2 - 10r + 3 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -10, c = 3 \):
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
Calculating the discriminant:
\[ r = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} \]
This gives us two possible values for \( r \):
1. \( r = \frac{18}{6} = 3 \)
2. \( r = \frac{2}{6} = \frac{1}{3} \)

### Step 6: Find the numbers
Now we can find the three numbers for both values of \( r \).

**Case 1: \( r = 3 \)**
- First number: \( \frac{6}{3} = 2 \)
- Second number: \( 6 \)
- Third number: \( 6 \cdot 3 = 18 \)

**Case 2: \( r = \frac{1}{3} \)**
- First number: \( \frac{6}{\frac{1}{3}} = 18 \)
- Second number: \( 6 \)
- Third number: \( 6 \cdot \frac{1}{3} = 2 \)

Thus, in both cases, the numbers are \( 2, 6, 18 \).

### Final Answer:
The three numbers are \( 2, 6, 18 \).

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