The sum of n terms of an A.P. is given by \(S_n = n^2 + n\).
We need to find the common difference, denoted by 'd'.
We know that the sum of the first n terms of an A.P. can also be represented as:
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
where 'a' is the first term and 'd' is the common difference.
We are given \(S_n = n^2 + n = n(n+1)\).
Let's find \(S_1\) and \(S_2\):
\(S_1 = 1^2 + 1 = 2\)
\(S_2 = 2^2 + 2 = 6\)
Since \(S_1\) represents the sum of the first term, it is equal to the first term itself:
\(a = S_1 = 2\)
\(S_2\) represents the sum of the first two terms, so:
\(S_2 = a + (a+d) = 2a + d = 6\)
We know \(a = 2\), so:
\(2(2) + d = 6\)
\(4 + d = 6\)
\(d = 6 - 4\)
\(d = 2\)
Therefore, the common difference is 2.
To solve this problem, we need to find the common difference of an arithmetic progression (A.P.) given that the sum of the first $ n $ terms, $ S_n $, is: $$ S_n = n^2 + n $$ Formula for Sum of First $ n $ Terms:
The sum of the first $ n $ terms of an arithmetic progression is given by the formula: $$ S_n = \frac{n}{2} \left( 2a + (n - 1) d \right) $$ Where: - $ a $ is the first term, - $ d $ is the common difference.
Difference of Consecutive Sums:
To find the common difference, we can calculate the difference between the sum of the first $ n $ terms and the sum of the first $ n-1 $ terms. This will give the $ n $-th term of the progression. $$ T_n = S_n - S_{n-1} $$ Substitute the given sum formula $ S_n = n^2 + n $ into this equation. $$ T_n = (n^2 + n) - ((n-1)^2 + (n-1)) $$ Simplifying this will help us find the common difference $ d $. Let me proceed with the simplification: $$ T_n = (n^2 + n) - ((n^2 - 2n + 1) + n - 1) $$ $$ T_n = n^2 + n - n^2 + 2n - 1 - n + 1 $$ $$ T_n = 2n $$ This expression tells us that the $ n $-th term is $ 2n $, and since the A.P. starts with 2 (i.e., $ T_1 = 2 $), we can conclude that the common difference is 2.
Answer: The common difference of the A.P. is 2, so the correct option is (C) 2.