Problem 35
Question
If 100 times the \(100^{\text {th }}\) term of an AP with non zero common difference equals the 50 times its \(50^{\text {th }}\) term, then the \(150^{\text {th }}\) term of this AP is : (a) \(-150\) (b) 150 times its \(50^{\text {th }}\) term (c) 150 (d) Zero
Step-by-Step Solution
Verified Answer
The 150th term of this AP is zero.
1Step 1: Understand the AP Formula
The nth term of an arithmetic progression (AP) can be expressed as \(a_n = a + (n-1) \cdot d\), where \(a\) is the first term and \(d\) is the common difference.
2Step 2: Set Up the Given Information
According to the problem, \(100 \times a_{100} = 50 \times a_{50}\). Using the formula for the nth term, this becomes \(100 \times (a + 99d) = 50 \times (a + 49d)\).
3Step 3: Simplify the Equation
Expand both sides of the equation: \(100a + 9900d = 50a + 2450d\). Then simplify it by moving all terms involving \(a\) and \(d\) to one side: \(100a - 50a + 9900d - 2450d = 0\).
4Step 4: Solve the Simplified Equation
Combine like terms: \(50a + 7450d = 0\). We can then express \(d\) in terms of \(a\): \(a = -149d\).
5Step 5: Find the 150th Term
Apply the nth term formula for the 150th term: \(a_{150} = a + 149d\). Substitute \(a = -149d\) into the equation: \(a_{150} = -149d + 149d = 0\).
Key Concepts
nth term formulacommon differencearithmetic sequence
nth term formula
An arithmetic progression (or arithmetic sequence) is a sequence of numbers in which the difference of any two successive members is a constant. This constant difference is known as the common difference, denoted by \( d \). To find any term in an arithmetic progression, we use a formula called the 'nth term formula'.
The formula for the nth term of an arithmetic progression is given by:
\[ a_n = a + (n-1) \cdot d \]
Here, \( a_n \) represents the nth term of the sequence, \( a \) is the first term, \( n \) is the position of the term in the sequence, and \( d \) is the common difference.
This formula is essential because it allows us to find any term in the sequence without having to list all the terms. For instance, if you need to find the 100th term, you'd substitute 100 for \( n \) in the formula. Understanding this relationship makes it easier to solve complex problems involving arithmetic sequences.
The formula for the nth term of an arithmetic progression is given by:
\[ a_n = a + (n-1) \cdot d \]
Here, \( a_n \) represents the nth term of the sequence, \( a \) is the first term, \( n \) is the position of the term in the sequence, and \( d \) is the common difference.
This formula is essential because it allows us to find any term in the sequence without having to list all the terms. For instance, if you need to find the 100th term, you'd substitute 100 for \( n \) in the formula. Understanding this relationship makes it easier to solve complex problems involving arithmetic sequences.
common difference
The common difference \( d \) in an arithmetic progression is what makes this type of sequence unique. It shows how each term in the sequence is related to the one before it, by adding or subtracting a fixed number each time.
To find the common difference \( d \), you can can simply subtract the first term from the second or the second from the third, and so on:
\[ d = a_2 - a_1 = a_3 - a_2 = \, \text{and so forth} \]
This constant defines how the sequence changes as it progresses. In our problem's solution, the common difference was used to relate the various terms to each other. Because the equation involved substituting terms of the form \( a + kd \), knowing the common difference was crucial for simplifying and ultimately solving the problem.
It's this characteristic of arithmetic progressions that helps us quickly identify patterns and solve problems like finding the nth term or verifying relationships between different terms.
To find the common difference \( d \), you can can simply subtract the first term from the second or the second from the third, and so on:
\[ d = a_2 - a_1 = a_3 - a_2 = \, \text{and so forth} \]
This constant defines how the sequence changes as it progresses. In our problem's solution, the common difference was used to relate the various terms to each other. Because the equation involved substituting terms of the form \( a + kd \), knowing the common difference was crucial for simplifying and ultimately solving the problem.
It's this characteristic of arithmetic progressions that helps us quickly identify patterns and solve problems like finding the nth term or verifying relationships between different terms.
arithmetic sequence
An arithmetic sequence is one where each term after the first is the sum of the previous term and a constant, the common difference \( d \). This type of sequence is straightforward and its pattern predictable, which makes arithmetic sequences very useful in various mathematical problems and real-life applications.
An arithmetic sequence can be represented with the notation:
\[ a, \, a+d, \, a+2d, \, a+3d, \, \ldots \]
In this representation, \( a \) is the first term and \( d \) is the common difference. This simple structure allows us to use the nth term formula to calculate any term's value directly.
In the problem given, the concepts of arithmetic sequence and its formula were used to find the terms' relationship and solve the equation. The sequence's consistency, through each term being separated by \( d \), made it feasible to arrive at an equation like \( 50a + 7450d = 0 \) and solve for \( d \) in terms of \( a \). The simplicity of arithmetic sequences is thus both a tool and a guide in understanding the nature and solutions of problems involving sequences.
An arithmetic sequence can be represented with the notation:
\[ a, \, a+d, \, a+2d, \, a+3d, \, \ldots \]
In this representation, \( a \) is the first term and \( d \) is the common difference. This simple structure allows us to use the nth term formula to calculate any term's value directly.
In the problem given, the concepts of arithmetic sequence and its formula were used to find the terms' relationship and solve the equation. The sequence's consistency, through each term being separated by \( d \), made it feasible to arrive at an equation like \( 50a + 7450d = 0 \) and solve for \( d \) in terms of \( a \). The simplicity of arithmetic sequences is thus both a tool and a guide in understanding the nature and solutions of problems involving sequences.
Other exercises in this chapter
Problem 33
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