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Chapter 10: Problem 50

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
The correct option is A: \(a = 2d\).

Step by step solution

01

Understand the Sum of an Arithmetic Progression

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

Substitute and Simplify Expression for \(S_{kx}\) and \(S_x\)

Using the formula for \(S_n\), substitute \(n = kx\) and \(n = x\) to find expressions for \(S_{kx}\) and \(S_x\):\[ S_{kx} = \frac{kx}{2} (2a + (kx-1)d), \]\[ S_x = \frac{x}{2} (2a + (x-1)d). \]
03

Find the Expression for \(\frac{S_{kx}}{S_x}\)

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

Determine the Independence from \(x\)

For \( \frac{S_{kx}}{S_x} \) to be independent of \(x\), the expression \( \frac{k(2a + (kx-1)d)}{2a + (x-1)d} \) must not have any terms involving \(x\) after simplification. Simplify:\[ k \left( \frac{2a + (kx-1)d}{2a + (x-1)d} \right) \text{ simplifies to be independent of } x, \] if the terms containing \(x\) in both numerator \( (kx-1)d \) and denominator \( (x-1)d \) cancel out, leading to:\[ 2a + kxd - d = 2a + xd - d \] or simply:\[ kxd - xd = 2a - 2a. \]By equating terms with \(x\): \( (k-1)xd = 0 \), leading us to infer that \( a = kd \).
05

Check Consistency with Options

Now verify which option matches our finding. We've simplified the independence condition to \( a = kd \), which for \(k = 2\) implies \(a = 2d\). Hence, the correct option is A: \(a = 2d\).

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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 where each term after the first is obtained by adding a constant value, known as the 'common difference.' To compute the sum of the first 'n' terms of an A.P., we use a specific formula. This formula helps us find the total when a series has a regular pattern. The sum of these terms, often denoted as \( S_n \), is given by the formula:\[S_n = \frac{n}{2} (2a + (n-1)d),\]where:
  • \( n \) is the number of terms,
  • \( a \) is the first term, and
  • \( d \) is the common difference between consecutive terms.
This formula is derived from the pattern that exists as terms repeat steps at equal intervals. It essentially calculates the average of the first and last terms (inside the parenthesis) and then multiplies by the number of terms. Understanding this formula is crucial for solving problems involving arithmetic series effectively.
Common Difference
The common difference in an arithmetic progression is the fixed amount added to each consecutive term of the sequence. Represented by \( d \), it determines how the sequence progresses.
  • If \( d \) is positive, each term increases as we move forward in the sequence.
  • If \( d \) is negative, each term decreases, leading to a descending series.
  • When \( d = 0 \), every term in the series is equal, making it a constant series.
Knowing \( d \) is crucial as it helps in predicting any term in the sequence using the formula:\[a_n = a + (n-1)d,\]where \( a_n \) is the nth term. This formula demonstrates how the sequence forms by repeatedly adding \( d \) to the first term. Understanding the common difference provides insights into the structure and growth pattern of the progression, helping to solve related problems easily.
First Term in Arithmetic Progression
In an arithmetic progression, the first term, denoted as \( a \), is the initial value from where the sequence starts. This term plays a pivotal role as it sets the baseline for all subsequent terms.
  • The choice of the first term affects the entire sequence, significantly influencing the values of the series.
  • It is used directly in the formula for both the sum and individual term expressions.
For example, to determine any term in an A.P., you use the formula:\[a_n = a + (n-1)d.\]Here, \( a \) ensures that as you add the product of the common difference and the term position, the progression follows its intended pattern accurately.In various exercises, determining the first term can often be the first step. It's essential for constructing the rest of the series and finding a specialized condition, as seen in problems where formulas must be independent of variables.

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Most popular questions from this chapter

For any odd integer \(n \geq 1\), \(n^{3}-(n-1)^{3}+\ldots+(-1)^{n-1} 1^{3}=\) (A) \(\frac{1}{2}(n-1)^{2}(2 n-1)\) (B) \(\frac{1}{4}(n-1)^{2}(2 n-1)\) (C) \(\frac{1}{2}(n+1)^{2}(2 n-1)\) (D) \(\frac{1}{4}(n+1)^{2}(2 n-1)\)

If \(a_{1}, a_{2}, \ldots, a_{n}\) are in A.P. with common difference \(d \neq 0\), then sum of the series \(\sin d\left[\sec a_{1} \sec a_{2}+\sec \right.\) \(\left.a_{2} \sec a_{3}+\ldots+\sec a_{n-1} \sec a_{n}\right]\) is (A) \(\tan a_{n}-\tan a_{1}\) (B) \(\cot a_{n}-\cot a_{1}\) (C) \(\sec a_{n}-\sec a_{1}\) (D) \(\operatorname{cosec} a_{n}-\operatorname{cosec} a_{1}\)

\(3^{1 / 3} \cdot 9^{1 / 9} \cdot 27^{1 / 27} \cdot 81^{1 / 81} \ldots\) upto \(\infty=\) (A) \(\sqrt{27}\) (B) \(\sqrt[3]{27}\) (C) \(\sqrt[4]{27}\) (D) None of these

If \(a, b, c, d\) and \(p\) are distinct real numbers such that \(\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2 p(a b+b c+c d)+\left(b^{2}+c^{2}+d^{2}\right)\) \(\leq 0\) then \(a, b, c, d\) are in (A) A.P. (B) G.P. (C) H.P. (D) \(a b=c d\)

The sum to \(n\) terms of the series \(\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\ldots\) is (A) \(n-\frac{\left(3^{n}-2^{n}\right)}{2^{n}}\) (B) \(n-\frac{2\left(3^{n}-2^{n}\right)}{3^{n}}\) (C) \(2^{n}-1\) (D) \(3^{n}-1\)
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