If Sr, denotes the sum of the first r terms of an AP then, \(\dfrac{S_{3r} - S_{r - 1}}{S_{2r} - S_{2r-1} }\)is
2r + 1
Understanding the sum of terms in an Arithmetic Progression (AP) is crucial for solving this problem. We are given an expression involving sums of different numbers of terms and asked to simplify it. Let the first term of the AP be \(a\) and the common difference be \(d\). The sum of the first \(n\) terms of an AP is given by the formula:
\[S_n = \dfrac{n}{2} [2a + (n-1)d]\]Also, the \(n\)-th term of an AP is given by:
\[a_n = a + (n-1)d\]A useful property relating the sum of terms and individual terms is that the difference between the sum of \(n\) terms and the sum of \(n-1\) terms gives the \(n\)-th term:
\[S_n - S_{n-1} = a_n\]Simplifying the Denominator of the Expression
The given expression is \(\dfrac{S_{3r} - S_{r - 1}}{S_{2r} - S_{2r-1} }\). Let's first simplify the denominator, \(S_{2r} - S_{2r-1}\). Using the property \(S_n - S_{n-1} = a_n\), where \(n = 2r\), we get:
\[S_{2r} - S_{2r-1} = a_{2r}\]The \(2r\)-th term of the AP is \(a_{2r} = a + (2r-1)d\).
So, the denominator simplifies to \(a + (2r-1)d\).
Simplifying the Numerator of the Expression
Now let's simplify the numerator, \(S_{3r} - S_{r-1}\). This represents the sum of terms from the \(r\)-th term up to the \(3r\)-th term, i.e., \(a_r + a_{r+1} + \dots + a_{3r}\).
The number of terms in this sequence is \((3r) - r + 1 = 2r + 1\).
This sequence of terms \(a_r, a_{r+1}, \dots, a_{3r}\) itself forms an Arithmetic Progression with:
- First term \(A = a_r = a + (r-1)d\)
- Last term \(L = a_{3r} = a + (3r-1)d\)
- Common difference \(D = d\)
- Number of terms \(N = 2r + 1\)
The sum of an AP can also be calculated using the formula \(S = \dfrac{N}{2}(A + L)\). Using this for the numerator:
\[S_{3r} - S_{r-1} = \dfrac{\text{Number of terms}}{2} (\text{First term} + \text{Last term})\] \[S_{3r} - S_{r-1} = \dfrac{2r+1}{2} [a_r + a_{3r}]\]Substitute the formulas for \(a_r\) and \(a_{3r}\):
\[S_{3r} - S_{r-1} = \dfrac{2r+1}{2} [(a + (r-1)d) + (a + (3r-1)d)]\] \[S_{3r} - S_{r-1} = \dfrac{2r+1}{2} [2a + (r-1 + 3r-1)d]\] \[S_{3r} - S_{r-1} = \dfrac{2r+1}{2} [2a + (4r-2)d]\] \[S_{3r} - S_{r-1} = \dfrac{2r+1}{2} [2a + 2(2r-1)d]\]Factor out 2 from the bracket:
\[S_{3r} - S_{r-1} = \dfrac{2r+1}{2} \cdot 2 [a + (2r-1)d]\] \[S_{3r} - S_{r-1} = (2r+1) [a + (2r-1)d]\]Evaluating the Entire Expression
Now substitute the simplified numerator and denominator back into the original expression:
\[\dfrac{S_{3r} - S_{r - 1}}{S_{2r} - S_{2r-1} } = \dfrac{(2r+1) [a + (2r-1)d]}{a + (2r-1)d}\]Assuming that the denominator \(a + (2r-1)d\) is not zero, we can cancel the term \([a + (2r-1)d]\) from both the numerator and the denominator.
\[\dfrac{S_{3r} - S_{r - 1}}{S_{2r} - S_{2r-1} } = 2r + 1\]Comparison with Options
Let's compare our simplified expression with the given options:
| Option | Value |
|---|---|
| 1 | 2r + 1 |
| 2 | 2r + 3 |
| 3 | 2r - 1 |
| 4 | 4r + 1 |
Our calculated value \(2r + 1\) matches Option 1.
Revision Table: Arithmetic Progression Concepts
| Concept | Formula/Explanation |
|---|---|
| \(n\)-th term of AP (\(a_n\)) | \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference. |
| Sum of first \(n\) terms of AP (\(S_n\)) | \(S_n = \dfrac{n}{2} [2a + (n-1)d]\) or \(S_n = \dfrac{n}{2} (a_1 + a_n)\) |
| Relationship between sum and term | \(S_n - S_{n-1} = a_n\) (for \(n > 1\)) |
| Sum of terms from \(a_k\) to \(a_m\) | Sum of an AP with first term \(a_k\), last term \(a_m\), and \((m-k+1)\) terms. |
Additional Information on AP Simplifications
This problem highlights how properties of APs can be used to simplify complex expressions involving sums. Recognizing that \(S_{2r} - S_{2r-1}\) is simply the \(2r\)-th term makes the denominator straightforward. For the numerator \(S_{3r} - S_{r-1}\), understanding that this represents the sum of terms from index \(r\) to \(3r\) allows us to treat this subsequence as a new AP and apply the sum formula. This method is often more efficient than expanding the sum formulas for \(S_{3r}\) and \(S_{r-1}\) and simplifying the resulting algebraic expression directly, which can be prone to errors.
The key steps were:
- Identify the structure of the expression involving sums of AP terms.
- Apply the property \(S_n - S_{n-1} = a_n\) to simplify the denominator.
- Recognize the numerator as the sum of a specific range of terms in the AP.
- Treat this range of terms as a new AP and apply the sum formula.
- Substitute the simplified forms back into the original expression and cancel common factors.
This systematic approach helps in solving problems involving sums of APs efficiently.
Important Questions from Arithmetic Progressions
What is \(\displaystyle \sum_{n=1}^{34} a_n\) equal to ?
The first and the second terms of an AP are \(\frac{5}{2}\) and \(\frac{23}{12}\) respectively. If nth term is the largest negative term, what is the value of n ?
In an AP, the first term is x and the sum of the first n terms is zero. What is the sum of next m terms ?
p, q, r and s are in AP such that p + s = 8 and qr = 15. What is the difference between largest and smallest numbers ?
The arithmetic mean of 1, 8, 27, 64, … up to n terms is given by
