A sequence is called an A.P if the difference of a term and the previous term is always same i.e if `a_(n+1)- a_(n)= ` constant ( common difference ) for all `n in N `
For an A.P whose first term is 'a ' and common difference is d has `n^(th)` term as `t_n=a+(n-1)d`
Sum of n terms of an A.P. whose first term is a, last term is `l` and common difference is d is
`S_(n) = n/2 (2a +(n-1)d)=n/2 (a+a+(n-1)d)= n/2 (a+l)`
If sum of n terms `S_n` for a sequence is given by `S_n=An^2+Bn+C`, then sequence is an A.P. whose common difference is

Options

A. A
B. B
C. 2A
D. 2B

Answer

To find the common difference of an arithmetic progression (A.P.) given that the sum of the first n terms \( S_n \) is expressed as \( S_n = An^2 + Bn + C \), we can follow these steps:

### Step 1: Understand the Sum of n Terms
The sum of the first n terms of an A.P. can be expressed in two ways:
1. \( S_n = \frac{n}{2} (2a + (n-1)d) \)
2. \( S_n = An^2 + Bn + C \)

### Step 2: Find the (n+1)th Term
To find the nth term \( t_n \) of the A.P., we can use the relationship:
\[ t_n = S_n - S_{n-1} \]

### Step 3: Calculate \( S_{n-1} \)
The sum of the first \( n-1 \) terms can be expressed as:
\[ S_{n-1} = A(n-1)^2 + B(n-1) + C \]
Expanding this gives:
\[ S_{n-1} = A(n^2 - 2n + 1) + B(n - 1) + C \]
\[ S_{n-1} = An^2 - 2An + A + Bn - B + C \]
\[ S_{n-1} = An^2 + (B - 2A)n + (A - B + C) \]

### Step 4: Find the nth Term \( t_n \)
Now, substituting \( S_n \) and \( S_{n-1} \) into the equation for \( t_n \):
\[ t_n = S_n - S_{n-1} \]
\[ t_n = (An^2 + Bn + C) - (An^2 + (B - 2A)n + (A - B + C)) \]
\[ t_n = Bn + C - (B - 2A)n - (A - B + C) \]
\[ t_n = (B - (B - 2A))n + (C - (A - B + C)) \]
\[ t_n = 2An + (B - A) \]

### Step 5: Find the (n-1)th Term \( t_{n-1} \)
Using the same method:
\[ t_{n-1} = S_{n-1} - S_{n-2} \]
Where \( S_{n-2} \) can be calculated similarly.

### Step 6: Calculate the Common Difference
The common difference \( d \) is given by:
\[ d = t_n - t_{n-1} \]
Substituting the expressions for \( t_n \) and \( t_{n-1} \):
\[ d = (2An + (B - A)) - (2A(n-1) + (B - A)) \]
\[ d = 2An + (B - A) - (2An - 2A + (B - A)) \]
\[ d = 2A \]

### Conclusion
Thus, the common difference \( d \) of the arithmetic progression is:
\[ d = 2A \]

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