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Chapter 8: Problem 49

If \(a\) is the first term, \(d\) the common difference and \(S_{k}\) the sum to \(k\) terms of an A.P., then for \(\frac{S_{k x}}{S_{x}}\) to be inde- pendent of \(x\) (A) \(a=2 d\) (B) \(a=d\) (C) \(2 a=d\) (D) None of these

Short Answer

Expert verified
For \(\frac{S_{kx}}{S_x}\) to be independent of \(x\), the condition \(2a = d\) (option C) must hold.

Step by step solution

01

Write the Formula for the Sum of an A.P.

For an arithmetic progression (A.P.) with first term \(a\) and common difference \(d\), the sum \(S_n\) of the first \(n\) terms is given by the formula: \[ S_n = \frac{n}{2} (2a + (n-1)d) \].
02

Calculate \(S_{kx}\) and \(S_x\)

First, find \(S_{kx}\):\[ S_{kx} = \frac{kx}{2} (2a + (kx-1)d) \].Next, find \(S_x\):\[ S_x = \frac{x}{2} (2a + (x-1)d) \].
03

Form the Ratio \(\frac{S_{kx}}{S_{x}}\)

The ratio \(\frac{S_{kx}}{S_{x}}\) is:\[ \frac{S_{kx}}{S_{x}} = \frac{\frac{kx}{2}(2a + (kx-1)d)}{\frac{x}{2}(2a + (x-1)d)} = \frac{k(2a + (kx-1)d)}{(2a + (x-1)d)} \].
04

Simplify the Ratio for Independence from \(x\)

For the ratio to be independent of \(x\), it must simplify to an expression not involving \(x\). This happens if the expressions \((kx-1)d\) and \((x-1)d\) cancel out, simplifying to a constant or polynomial independent of \(x\). Simplifying gives:\[ k(2a + kxd - d) = 2a + xd - d \].Solving for conditions that make the \(x\) terms cancel out, solving \(kxd - xd = 2a(1-k) + d(1-k) = 0\), leads to \(2a = d\).
05

Verify the Correct Condition

Putting \(2a = d\) back into the expression confirms that \(kx = x\), simplifying \(k\) out as \((2a - d) = 0\) becomes a constant. Hence \(a = \frac{d}{2}\) ensures independence from \(x\), making option (C) correct.

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Key Concepts

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

Sum of Arithmetic Progression
An arithmetic progression (A.P.) is a sequence of numbers in which the difference between consecutive terms is the same. When calculating the sum of an A.P., it's essential to have a clear formula in mind. This formula provides a structured way to sum a series of numbers in a sequence.

The sum of the first \(n\) terms of an A.P. can be determined using the formula: \[S_n = \frac{n}{2} (2a + (n-1)d)\] where:
  • \(S_n\) is the sum of the first \(n\) terms.
  • \(a\) is the first term of the sequence.
  • \(d\) is the common difference between consecutive terms.
This formula essentially works by averaging the first and the last term of the series and multiplying by the number of terms \(n\). It's crucial to understand this principle as it applies to various types of problems involving sequence sums.
Common Difference
The common difference \(d\) in an arithmetic progression determines how each term is spaced from the next. It is a constant value, meaning every number in the sequence is separated by \(d\). This uniform spacing is what defines an arithmetic sequence.

In the formula \(S_n = \frac{n}{2} (2a + (n-1)d)\), the term \((n-1)d\) shows how the common difference affects the sum of multiple terms. This part of the formula accounts for the incremental buildup in the sum due to adding \(d\) after each term.

The importance of the common difference extends beyond calculation. It provides insight into how sequences behave. Recognizing the common difference helps identify the type of progression, predict future terms, and understand trends. Mastery of this concept is key in grasping broader mathematical sequences.
Independence of Variable
Independence of a variable in mathematical expressions ensures that the outcome doesn't change with that specific variable. In the context of the problem, the goal was to make the expression \(\frac{S_{kx}}{S_{x}}\) independent of \(x\).

For this independence, terms involving \(x\) should cancel out, so the result is unaffected by changes in \(x\). This characteristic is achieved when we ensure that certain conditions are met. Specifically, if \(2a = d\), it turns the expression into a result that doesn't rely on \(x\).

This concept is crucial because independence from variables often simplifies complex expressions, allowing for easier manipulation and understanding. It emphasizes the conditions needed to transform variable-dependent scenarios into constant or predictable outcomes.

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Most popular questions from this chapter

If \(a, b, c\) are in A.P. and \(a^{2}, b^{2}, c^{2}\) arc in H.P. then (A) \(a=b=c\) (B) \(-\frac{a}{2}, b, c\) are in G.P. (C) \(-\frac{c}{2}, b, a\) are in G.P. (D) \(-\frac{a}{2}, b, c\) are in H.P.

Let \(\left(1+x^{2}\right)^{2}(1+x)^{n}=\sum_{k=0} a_{k} x^{k} .\) If \(a_{1}, a_{2}, a_{3}\), are in A.P., then \(n\) is equal to (A) 1 (B) 2 (C) 3 (D) 4

Let \(a_{1}, a_{2}, a_{3}, \ldots\) be terms of an A.P. If \(\frac{a_{1}+a_{2}+\ldots a_{p}}{a_{1}+a_{2} \ldots+a_{q}}\) \(=\frac{p^{2}}{q^{2}}, p \neq q\), then \(\frac{a_{6}}{a_{21}}\) equals \([2006]\) (A) \(\frac{41}{11}\) (B) \(\frac{7}{2}\) (C) \(\frac{2}{7}\) (D) \(\frac{11}{41}\)

The sum of first \(n\) terms of the series \(1 \cdot 1 !+2 \cdot 2 !+3 \cdot 3 !+4 \cdot 4 !+\ldots\) is (A) \((n+1) !-1\) (B) \(n !-1\) (C) \((n-1) !-1\) (D) None of these

\(x=\sum_{n=0}^{\infty} a^{n}, y=\sum_{n=0}^{\infty} b^{n}, z=\sum_{n=0}^{\infty} c^{n}\) where \(a, b, c\) are in A.P. and \(|a|<1,|b|<1,|c|<1\), then \(x, y, z\) are in \([2005]\) (A) G.P. (B) A.P. (C) Arithmetic - Geometric Progression (D) H.P.
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