Sum of 8 consecutive even number is `S_(1).`
Quantity I- Sum of second number and eighth number in `S_(1)`
Quantity II- Sum of third number sixth number in `S_(1)`

Options

A. Quantity `Igt` Quantity II
B. Quantity `I lt` Quantity II
C. Quantity `I ge` Quantity II
D. Quantity `I le` Quantity II

Explanation

To solve the problem step by step, we will first define the 8 consecutive even numbers and then calculate the required quantities.

### Step 1: Define the 8 consecutive even numbers
Let's assume the first even number is \( X \). Then the 8 consecutive even numbers can be represented as:
- 1st number: \( X \)
- 2nd number: \( X + 2 \)
- 3rd number: \( X + 4 \)
- 4th number: \( X + 6 \)
- 5th number: \( X + 8 \)
- 6th number: \( X + 10 \)
- 7th number: \( X + 12 \)
- 8th number: \( X + 14 \)

### Step 2: Calculate the sum \( S_1 \)
The sum \( S_1 \) of these 8 consecutive even numbers is:
\[ S_1 = X + (X + 2) + (X + 4) + (X + 6) + (X + 8) + (X + 10) + (X + 12) + (X + 14) \]
This simplifies to:
\[ S_1 = 8X + (2 + 4 + 6 + 8 + 10 + 12 + 14) \]
Calculating the sum of the constants:
\[ 2 + 4 + 6 + 8 + 10 + 12 + 14 = 56 \]
Thus, we have:
\[ S_1 = 8X + 56 \]

### Step 3: Calculate Quantity I
Quantity I is the sum of the 2nd number and the 8th number:
\[ \text{2nd number} = X + 2 \]
\[ \text{8th number} = X + 14 \]
Now, summing these:
\[ \text{Quantity I} = (X + 2) + (X + 14) = 2X + 16 \]

### Step 4: Calculate Quantity II
Quantity II is the sum of the 3rd number and the 6th number:
\[ \text{3rd number} = X + 4 \]
\[ \text{6th number} = X + 10 \]
Now, summing these:
\[ \text{Quantity II} = (X + 4) + (X + 10) = 2X + 14 \]

### Step 5: Compare Quantity I and Quantity II
Now we compare the two quantities:
- Quantity I: \( 2X + 16 \)
- Quantity II: \( 2X + 14 \)

Since both quantities have the same \( 2X \) term, we can compare the constants:
\[ 2X + 16 > 2X + 14 \]
Thus, Quantity I is greater than Quantity II.

### Conclusion
The answer is that Quantity I is greater than Quantity II.

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