Let `s_(n)` be the sum of the first n terms of an AP. If `S_(2n)=3n+14n^(2)`, then what is the common difference ?

Options

A. A)5
B. B)6
C. C)7
D. D)9

Answer

To solve the problem, we need to find the common difference of an arithmetic progression (AP) given that the sum of the first \(2n\) terms is \(S(2n) = 3n + 14n^2\).

### Step 1: Understand the formula for the sum of the first \(n\) terms of an AP
The sum of the first \(n\) terms of an AP can be expressed as:
\[ S(n) = \frac{n}{2} \times (2a + (n-1)d) \]
where \(a\) is the first term and \(d\) is the common difference.

### Step 2: Express \(S(2n)\)
Using the formula, we can express \(S(2n)\):
\[ S(2n) = \frac{2n}{2} \times (2a + (2n-1)d) = n(2a + (2n-1)d) \]
Given that \(S(2n) = 3n + 14n^2\), we can equate:
\[ n(2a + (2n-1)d) = 3n + 14n^2 \]

### Step 3: Simplify the equation
Dividing both sides by \(n\) (assuming \(n \neq 0\)):
\[ 2a + (2n-1)d = 3 + 14n \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ (2n-1)d = 14n + 3 - 2a \]
This equation will help us find the common difference \(d\).

### Step 5: Calculate \(S(2)\) and \(S(4)\)
To find \(d\), we will calculate \(S(2)\) and \(S(4)\):
- For \(n = 1\):
\[ S(2) = 3(1) + 14(1)^2 = 3 + 14 = 17 \]
- For \(n = 2\):
\[ S(4) = 3(2) + 14(2)^2 = 6 + 56 = 62 \]

### Step 6: Use the values of \(S(2)\) and \(S(4)\) to find \(d\)
Using the formula \(d = \frac{S(4) - 2S(2)}{4}\):
\[ d = \frac{62 - 2 \times 17}{4} \]
Calculating this gives:
\[ d = \frac{62 - 34}{4} = \frac{28}{4} = 7 \]

### Conclusion
The common difference \(d\) is \(7\).