`a+ar+ar^2+ . . . + oo=15`
`a^2+(ar)^2+(ar^2)^2+ . . . + oo=150` .
Find `ar^2+ar^4+ar^6+. . . oo`

Options

A. `1/2`
B. `2/5`
C. `1/6`
D. None

Answer

To solve the problem, we start with the two given equations:

1. \( S_1 = a + ar + ar^2 + \ldots = 15 \)
2. \( S_2 = a^2 + (ar)^2 + (ar^2)^2 + \ldots = 150 \)

### Step 1: Express the sums in terms of \( a \) and \( r \)

The sum of an infinite geometric series can be expressed as:

\[ S = \frac{a}{1 - r} \]

For the first equation:

\[ \frac{a}{1 - r} = 15 \tag{1} \]

For the second equation, we note that:

\[ S_2 = a^2 + (ar)^2 + (ar^2)^2 + \ldots = \frac{a^2}{1 - r^2} \]

Thus, we have:

\[ \frac{a^2}{1 - r^2} = 150 \tag{2} \]

### Step 2: Solve for \( a \) and \( r \)

From equation (1), we can express \( a \) in terms of \( r \):

\[ a = 15(1 - r) \tag{3} \]

Substituting equation (3) into equation (2):

\[ \frac{(15(1 - r))^2}{1 - r^2} = 150 \]

Expanding this gives:

\[ \frac{225(1 - r)^2}{1 - r^2} = 150 \]

Cross-multiplying yields:

\[ 225(1 - r)^2 = 150(1 - r^2) \]

Expanding both sides:

\[ 225(1 - 2r + r^2) = 150(1 - r^2) \]

This simplifies to:

\[ 225 - 450r + 225r^2 = 150 - 150r^2 \]

Combining like terms:

\[ 375r^2 - 450r + 75 = 0 \]

### Step 3: Simplify the quadratic equation

Dividing the entire equation by 75:

\[ 5r^2 - 6r + 1 = 0 \]

### Step 4: Use the quadratic formula to find \( r \)

Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ r = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 5 \cdot 1}}{2 \cdot 5} \]

Calculating the discriminant:

\[ r = \frac{6 \pm \sqrt{36 - 20}}{10} = \frac{6 \pm \sqrt{16}}{10} = \frac{6 \pm 4}{10} \]

Thus, we have two potential solutions for \( r \):

\[ r = \frac{10}{10} = 1 \quad \text{(not valid for a converging series)} \]
\[ r = \frac{2}{10} = \frac{1}{5} \]

### Step 5: Find \( a \)

Substituting \( r = \frac{1}{5} \) back into equation (3):

\[ a = 15(1 - \frac{1}{5}) = 15 \cdot \frac{4}{5} = 12 \]

### Step 6: Find \( ar^2 + ar^4 + ar^6 + \ldots \)

This series can be expressed as:

\[ ar^2 + ar^4 + ar^6 + \ldots = ar^2(1 + r^2 + r^4 + \ldots) \]

The sum of the infinite series \( 1 + r^2 + r^4 + \ldots \) is:

\[ \frac{1}{1 - r^2} \]

Thus, we have:

\[ ar^2 \cdot \frac{1}{1 - r^2} \]

Calculating \( ar^2 \):

\[ ar^2 = 12 \cdot \left(\frac{1}{5}\right)^2 = 12 \cdot \frac{1}{25} = \frac{12}{25} \]

Now substituting \( r^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25} \):

\[ 1 - r^2 = 1 - \frac{1}{25} = \frac{24}{25} \]

Thus, the sum becomes:

\[ \frac{\frac{12}{25}}{\frac{24}{25}} = \frac{12}{24} = \frac{1}{2} \]

### Final Answer

The value of \( ar^2 + ar^4 + ar^6 + \ldots \) is:

\[ \boxed{\frac{1}{2}} \]