Problem 21
Question
Find the nth term of the arithmetic sequence \(\left\\{a_{n}\right\\}\) whose first term \(a_{1}\) and common difference d are given. What is the 51st term? $$ a_{1}=0 ; \quad d=\frac{1}{2} $$
Step-by-Step Solution
Verified Answer
The 51st term is 25.
1Step 1: Understand the Formula for the nth Term of an Arithmetic Sequence
The nth term of an arithmetic sequence can be found using the formula: \[ a_{n} = a_{1} + (n - 1) \times d \]where \(a_n\) is the nth term, \(a_{1}\) is the first term, \(d\) is the common difference, and \(n\) is the term number.
2Step 2: Substitute the Given Values
Substitute the given values into the formula. We know that \(a_{1} = 0\) and \(d = \frac{1}{2}\). We are looking for the 51st term, so \(n = 51\). Plugging in these values, we get: \[ a_{51} = 0 + (51 - 1) \times \frac{1}{2} \]
3Step 3: Simplify the Expression
First calculate the expression inside the parentheses:\[ 51 - 1 = 50 \]Now multiply by the common difference \( \frac{1}{2} \):\[ 50 \times \frac{1}{2} = 25 \]Thus, the 51st term is \[ a_{51} = 25 \]
Key Concepts
nth termcommon differencearithmetic progressionsequence formula
nth term
In any arithmetic sequence, the 'nth term' refers to any term in the sequence denoted by the position number 'n'. The formula to find the nth term is very straightforward: \ \ \( a_{n} = a_{1} + (n - 1) \times d \) \ \ Here: \ \ - \(a_{n}\) is the nth term \ - \(a_{1}\) is the first term \ - \(d\) is the common difference \ - \(n\) is the term number \ \ The formula is used to find any term in the sequence. It uses the first term as a starting point and the common difference to calculate how much to add to move between terms.
common difference
The 'common difference' is a key element in an arithmetic sequence. It’s the amount added or subtracted from one term to the next. In the sequence we're dealing with, the common difference (d) is \( \frac{1}{2} \), meaning each term is half more than the previous term. \ \ - If the sequence were \{0, 0.5, 1, 1.5, 2, ...\}, \( \frac{1}{2} \) would be the difference between each term. \ - The formula for the nth term incorporates the common difference directly: \( a_{n} = a_{1} + (n-1) \times d \) \ \ Without the common difference, finding any term would be difficult because you'd have no regular change to count on.
arithmetic progression
An 'arithmetic progression' (or arithmetic sequence) is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is known as the 'common difference'. \ \ For our arithmetic sequence where \(a_1 = 0 \) and \(d = \frac{1}{2} \): \ \ - The progression would look like \(0, 0.5, 1, 1.5, 2, ...\) \ - Each number in the sequence is found by adding a constant value (\(\frac{1}{2}\) in this case) to the previous term. \ \ The arithmetic progression can be visually represented as regularly spaced points on a number line or graph.
sequence formula
The 'sequence formula' for an arithmetic sequence allows you to find any term in that sequence directly. The general formula is: \( a_{n} = a_{1} + (n-1) \times d \). This formula eliminates the need to list all the previous terms to find a specific term. \ \ For example, with \( a_{1} = 0 \) and \( d = \frac{1}{2} \), to find the 51st term (\(a_{51}\)): \ \ \( a_{51} = 0 + (51-1) \times \frac{1}{2} \) \ \ \( a_{51} = 0 + 50 \times \frac{1}{2} \) \ \ \( a_{51} = 25 \) \ \ It's a quick way to find any term without counting each one.
Other exercises in this chapter
Problem 21
Find the fifth term and the nth term of the geometric sequence whose first term \(a_{1}\) and common ratio \(r\) are given. $$ a_{1}=5 ; \quad r=-1 $$
View solution Problem 21
List the first five terms of each sequence. \(\left\\{s_{n}\right\\}=\left\\{\frac{3^{n}}{2^{n}+3}\right\\}\)
View solution Problem 22
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ n(n+1)(n+2) \text { is divisible by } 6 $
View solution Problem 22
Expand each expression using the Binomial Theorem. $$ (2 x+3)^{5} $$
View solution