/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Let \(a_{1}, a_{2}, \ldots ., a_... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 4

Let \(a_{1}, a_{2}, \ldots ., a_{n}\) be a given A.P. whose common difference is an integer and \(S_{n}=a_{1}+a_{2}+\ldots+a_{n} .\) If \(a_{1}=1, a_{n}=300\) and \(15 \leq n \leq 50\), then the ordered pair \(\left(S_{n-4}, a_{n-4}\right)\) is equal to : (a) \((2490,249)\) (b) \((2480,249)\) (c) \((2480,248)\) (d) \((2490,248)\)

Short Answer

Expert verified
The ordered pair is \((2490, 248)\).

Step by step solution

01

Understand the arithmetic progression (AP)

Given that we have an arithmetic progression (AP) with the first term as \(a_1 = 1\), common difference \(d\), and the \(n\)-th term \(a_n = 300\), we need to find \(S_{n-4}\) and \(a_{n-4}\). The number of terms, \(n\), is such that \(15 \leq n \leq 50\).
02

Determine the common difference

The \(n\)-th term of an AP is given by the formula \(a_n = a_1 + (n-1)d\). Substitute the known values: \[300 = 1 + (n - 1)d\]Simplify to find:\[299 = (n - 1)d\]
03

Consider the integer factors of 299

Since \(d\) is an integer, solve for \(d\) by dividing 299 by possible values of \(n - 1\) such that \(d\) is an integer. Recall that \(15 \leq n \leq 50\) translates to \(14 \leq n-1 \leq 49\). Test each factor: 299 can be divided by 1, 13, 23, and 299.
04

Calculate for possible values of \(n\)

To ensure \(15 \leq n \leq 50\), check valid combinations.1. \(n-1 = 23\), then \(d = 13\), so \(n = 24\), which is valid.
05

Calculate \(S_{n-4}\)

Since \(n = 24\), \(n-4 = 20\). The sum of the first \(m\) terms of an AP is:\[S_m = \frac{m}{2} \times (2a_1 + (m-1)d)\]So for \(m = 20\),\[S_{20} = \frac{20}{2} \times (2 \times 1 + (20-1) \times 13)\]\[S_{20} = 10 \times (2 + 247) = 10 \times 249 = 2490\]
06

Calculate \(a_{n-4}\)

The term \(a_{m}\) in an AP is given by:\[a_{m} = a_1 + (m-1)d\]So for \(m = 20\),\[a_{20} = 1 + (20 - 1) \times 13 = 1 + 247 = 248\]
07

Determine the ordered pair \((S_{n-4}, a_{n-4})\)

From the calculations:- \(S_{n-4} = 2490\)- \(a_{n-4} = 248\)Therefore, the ordered pair is \((2490, 248)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Difference
In arithmetic progressions (AP), the common difference is a vital feature that defines the sequence. It is the difference between consecutive terms, making each term predictable. For example, if the first term is 1 and the common difference is 13, the sequence reads: 1, 14, 27, 40, and so on. This consistency allows for patterns in an AP that aid in problem-solving.
To calculate it mathematically, use the formula for the n-th term of an AP: \[\ a_n = a_1 + (n-1) \cdot d\ \] where:
  • \(a_n\) is the nth term.
  • \(a_1\) is the first term.
  • \(d\) is the common difference.
  • \(n\) is the term number you are targeting.
By rearranging this formula, you can find the common difference when you know two specific terms in the sequence. This helps estimate the next terms or recalculate previous terms, making it an invaluable asset when navigating APs.
Sum of Terms
The sum of the terms in an arithmetic progression can be determined using a specific formula. This is crucial for summing a series of numbers efficiently without adding them one by one. For calculating the sum of the first \(m\) terms, use:\[\ S_m = \frac{m}{2} \times (2a_1 + (m-1)d)\ \]Where:
  • \(S_m\) is the sum of the first \(m\) terms.
  • \(m\) is the number of terms being summed.
  • \(a_1\) is the first term.
  • \(d\) is the common difference.
Using this formula, you can compute large sums swiftly. For example, if the sum of the first 20 terms needs to be calculated and the sequence starts at 1 with a common difference of 13, you can find:\[\ S_{20} = 10 \times (2 + 247) = 2490\ \] thus confirming our earlier computation. Recognizing this formula allows you to tackle extensive arithmetic series without straightforwardly relying on intuitive methods.
Sequence Analysis
Analyzing an arithmetic progression involves understanding how the sequence behaves, changes, and is formed over a set of terms. It starts from knowing the individual components, such as the common difference and each term's calculation.Here鈥檚 how:- **Determine the first term and how each subsequent term builds off the previous.** Knowing the initial term in an AP sets the foundation from which all other terms are developed.- **Understand the relationship between terms through the common difference.** This aids in analyzing how quickly or slowly the sequence grows.- **Determine sum properties and specific term positioning.** Use the formulas to find specific terms (like \(a_{n-4}\) in this scenario) and sum totals like \(S_{n-4}\). By finding \(a_{20} = 248\) when starting at 1 and moving by 13, the comprehension of sequence positioning informs the entire sequence structure.- **Apply these understandings to calculate efficiently and solve predictably.** This involves blending sequence understanding with mathematical precision through tailored formulas provided in arithmetic progressions to streamline any calculation or sequence probing.

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Most popular questions from this chapter

If \(1+\left(1-2^{2} \cdot 1\right)+\left(1-4^{2} \cdot 3\right)+\left(1-6^{2} \cdot 5\right)+\ldots \ldots+\left(1-20^{2},\right.\), 19) \(=\alpha-220 \beta\), then an ordered pair \((\alpha, \beta)\) is equal to: [Sep. 04, 2020 (I)] (a) \((10,97)\) (b) \((11,103)\) (c) \((10,103)\) (d) \((11,97)\)

The least positive integer \(\mathrm{n}\) such that \(1-\frac{2}{3}-\frac{2}{3^{2}}-\ldots .-\frac{2}{3^{n-1}}<\frac{1}{100}\), is: [Online April 12, 2014] (a) \(\underline{4}\) (b) 5 (c) 6 (d) 7

(a) 81 (b) \(-127\) (c) \(-81\) (d) 127 If \(3^{2 \sin 2 \alpha-1}, 14\) and \(3^{4-2 \sin 2 \alpha}\) are the first three terms of an A.P. for some \(\alpha\), then the sixth term of this A.P is: [Sep. \(05,2020(\mathrm{I})]\) (a) 66 (b) 81 (c) 65 (d) 78

The sum of all natural numbers ' \(n\) ' such that \(1001\) is: [April 08, 2019 (I)] (a) 3203 (b) 3303 (c) 3221 (d) 3121

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243 , then the sum of the first 50 terms of this G.P. is: [Sep. \(05, \mathbf{2 0 2 0}\) (II)] (a) \(\frac{1}{26}\left(3^{49}-1\right)\) (b) \(\frac{1}{26}\left(3^{50}-1\right)\) (c) \(\frac{2}{13}\left(3^{50}-1\right)\) (d) \(\frac{1}{13}\left(3^{50}-1\right)\)
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