Let S be the sum of the first n terms of the arithmetic sequence 3, 7, 11, ...., and let T be the sum of the first n terms of the arithmetic sequence 8 , 10 , 12 ,.... For n `gt` 1 , S = T for

Options

A. no value of n
B. one value of n
C. two values of n
D. three values of n

Explanation

To solve the problem, we need to find the value of \( n \) such that the sum of the first \( n \) terms of the two given arithmetic sequences are equal.

### Step 1: Identify the sequences and their parameters
The first arithmetic sequence is:
- First term \( a_1 = 3 \)
- Common difference \( d_1 = 7 - 3 = 4 \)

The second arithmetic sequence is:
- First term \( a_2 = 8 \)
- Common difference \( d_2 = 10 - 8 = 2 \)

### Step 2: Write the formula for the sum of the first \( n \) terms of an arithmetic sequence
The sum \( S \) of the first \( n \) terms of an arithmetic sequence can be calculated using the formula:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \]

### Step 3: Calculate the sum \( S \) for the first sequence
For the first sequence:
\[ S = \frac{n}{2} \left( 2 \times 3 + (n-1) \times 4 \right) \]
Simplifying this:
\[ S = \frac{n}{2} \left( 6 + 4n - 4 \right) = \frac{n}{2} \left( 4n + 2 \right) = n(2n + 1) \]

### Step 4: Calculate the sum \( T \) for the second sequence
For the second sequence:
\[ T = \frac{n}{2} \left( 2 \times 8 + (n-1) \times 2 \right) \]
Simplifying this:
\[ T = \frac{n}{2} \left( 16 + 2n - 2 \right) = \frac{n}{2} \left( 2n + 14 \right) = n(n + 7) \]

### Step 5: Set the two sums equal to each other
We set \( S = T \):
\[ n(2n + 1) = n(n + 7) \]

### Step 6: Simplify the equation
Assuming \( n \neq 0 \) (since \( n > 1 \)):
\[ 2n + 1 = n + 7 \]
Subtract \( n \) from both sides:
\[ n + 1 = 7 \]
Subtract 1 from both sides:
\[ n = 6 \]

### Step 7: Conclusion
The only value of \( n \) for which the sums are equal is \( n = 6 \).