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.
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: