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
Finding the Common Difference of an AP from the Sum Formula
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.
Understanding the Sum of an AP Formula
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:
\(S_n = \frac{n}{2}(2a) + \frac{n}{2}(n-1)d\)
\(S_n = na + \frac{n(n-1)}{2}d\)
Comparing the Given Formula with the Standard Formula
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 (\(d\)).
- The coefficient of \(n\) in the given formula is \(P\). In the standard formula, it is \(a\). Therefore, the first term \(a = P\).
- The coefficient of \(\frac{n(n-1)}{2}\) in the given formula is \(Q\). In the standard formula, it is \(d\). Therefore, the common difference \(d = Q\).
Alternative Method: Using the First Few Terms
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 \(Q\).
Conclusion on Common Difference
Based on the analysis of the given sum formula for the AP, the common difference is \(Q\).
Let's look at the options:
- Option 1: \(P + Q\)
- Option 2: \(2P + 3Q\)
- Option 3: \(2Q\)
- Option 4: \(Q\)
Our calculated common difference matches Option 4.
Revision Table: AP Concepts
| 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 \(n > 1\)) |
Additional Information: Properties of AP
An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by \(d\).
- The general form of an AP is \(a, a+d, a+2d, a+3d, \dots\)
- The \(n\)th term of an AP can be found if the first term and common difference are known.
- The sum of an AP can be calculated using formulas involving the first term, common difference, and the number of terms.
- The formula for \(S_n\) can be written as \(S_n = (\text{constant}_1)n + (\text{constant}_2)n^2\). The given formula \(S_n = nP + \frac{Q}{2}n(n-1) = nP + \frac{Q}{2}n^2 - \frac{Q}{2}n = (P - \frac{Q}{2})n + \frac{Q}{2}n^2\) is in this form.
- For an AP, if \(S_n = An^2 + Bn\), the common difference \(d = 2A\). In the given formula, \(A = Q/2\), so \(d = 2(Q/2) = Q\). The first term \(a = A + B\). In the given formula, \(B = P - Q/2\), so \(a = Q/2 + (P - Q/2) = P\). This confirms our earlier results.
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