If \({S_n} = nP + \frac{{n\left( {n - 1} \right)Q}}{2}\) , where S ndenotes the sum of the first n terms of an AP, then the common difference is
Q
The question provides a formula for the sum of the first \(n\) terms of an Arithmetic Progression (AP), denoted as \(S_n\). The given formula is:
\({S_n} = nP + \frac{{n\left( {n - 1} \right)Q}}{2}\)
We need to find the common difference of this AP.
The standard formula for the sum of the first \(n\) terms of an AP with the first term \(a\) and common difference \(d\) is:
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
This formula can be expanded as:
\(S_n = \frac{n}{2}(2a) + \frac{n}{2}(n-1)d\)
\(S_n = na + \frac{n(n-1)}{2}d\)
We can compare the given formula \({S_n} = nP + \frac{{n\left( {n - 1} \right)Q}}{2}\) with the standard form \(S_n = na + \frac{n(n-1)}{2}d\).
By comparing the coefficients of \(n\) and \(\frac{n(n-1)}{2}\), we can identify the first term (\(a\)) and the common difference (\(d\)).
We can also find the common difference by calculating the first two terms of the AP using the given \(S_n\) formula.
The first term, \(a_1\), is equal to the sum of the first term, \(S_1\).
\(S_1 = (1)P + \frac{{1\left( {1 - 1} \right)Q}}{2}\)
\(S_1 = P + \frac{{1 \times 0 \times Q}}{2}\)
\(S_1 = P + 0 = P\)
So, the first term \(a_1 = P\).
The sum of the first two terms, \(S_2\), is equal to the sum of the first term (\(a_1\)) and the second term (\(a_2\)).
\(S_2 = a_1 + a_2\)
Using the given \(S_n\) formula for \(n=2\):
\(S_2 = (2)P + \frac{{2\left( {2 - 1} \right)Q}}{2}\)
\(S_2 = 2P + \frac{{2 \times 1 \times Q}}{2}\)
\(S_2 = 2P + Q\)
Now, we know that \(S_2 = a_1 + a_2\). Substitute the values of \(S_2\) and \(a_1\):
\(2P + Q = P + a_2\)
To find \(a_2\), subtract \(P\) from both sides:
\(a_2 = (2P + Q) - P\)
\(a_2 = P + Q\)
The common difference \(d\) of an AP is the difference between any term and its preceding term. For example, \(d = a_2 - a_1\).
\(d = (P + Q) - P\)
\(d = Q\)
Both methods show that the common difference of the AP is \(Q\).
Based on the analysis of the given sum formula for the AP, the common difference is \(Q\).
Let's look at the options:
Our calculated common difference matches Option 4.
| Concept | Formula/Description |
|---|---|
| nth term of AP (\(a_n\)) | \(a_n = a + (n-1)d\) |
| Sum of first n terms (\(S_n\)) | \(S_n = \frac{n}{2}[2a + (n-1)d]\) or \(S_n = \frac{n}{2}(a_1 + a_n)\) |
| Common Difference (\(d\)) | \(d = a_n - a_{n-1}\) (for \(n > 1\)) |
| Relation between \(S_n\), \(S_{n-1}\), and \(a_n\) | \(a_n = S_n - S_{n-1}\) (for \(n > 1\)) |
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