Answer:
[tex]Q 2: \quad a_{18} = -52[/tex]
[tex]Q3:\quad a_{50} = \dfrac{37}{3} = 12 \dfrac{1}{3}[/tex]
[tex]Q4:\quad a_{2023} = 4043[/tex]
Step-by-step explanation:
(I am not quite sure why you mentioned that some cannot be solved by the arithmetic sequence formula. Questions 2 - 4 are directly solvable by the arithmetic sequence formula)
First let's examine the sample problem and understand what the various terms mean
The formula for an arithmetic sequence is
[tex]a_n = a_1 + (n - 1)d\\\\$where\\a_n = $ nth term\\a_1 = $ first term\\n = number of terms\\d = common difference = difference between successive terms[/tex]
Question 2:
[tex]d = -5, a_1 = 33, n = 18[/tex]
[tex]a_{18} = 33+(18-1)(-5)\\\\a_{18} = 33 + (17)(-5)\\\\a_{18} = 33 - 85\\\\a_{18} = -52\\\\[/tex]
Question 3:
[tex]d = \dfrac{1}{3} - \dfrac{1}{12} = = \dfrac{4}{12} - \dfrac{1}{12} = \dfrac{3}{12} =\dfrac{1}{4}[/tex]
[tex]a_1 = \dfrac{1}{12}[/tex]
[tex]a_{50} = \dfrac{1}{12} + (50-1)\dfrac{1}{4}\\\\a_{50} = \dfrac{1}{12} + 49 \cdot \dfrac{1}{4}\\\\a_{50} = \dfrac{1}{12}+ \dfrac{49}{4}\\\\ a_{50} = \dfrac{1}{12}+ \dfrac{147}{12}\\\\ a_{50} = \dfrac{148}{12}\\\\ a_{50} = \dfrac{37}{3}\\\\a_{50} = 12 \dfrac{1}{3}[/tex]
Question 4
[tex]d = 3, a_1 = -2023, n = 2023[/tex]
[tex]a_{2023} = -2023 + (2023-1)(3)\\\\a_{2023} = -2023 + 2022 (3)\\\\a_{2023} = -2023 + 6066\\\\a_{2023} = 4043[/tex]
I hope that is what you were looking for
Answer:
[tex]\textsf{2.}\quad a_n=38-5n\: ; \quad a_{18}=-52[/tex]
[tex]\textsf{3.}\quad a_n=\dfrac{1}{4}n-\dfrac{1}{6}\: ; \quad a_{50}=\dfrac{37}{3}[/tex]
[tex]\textsf{4.}\quad a_n=3n-2026\: ; \quad a_{2023}=4043[/tex]
Step-by-step explanation:
An arithmetic sequence is a sequence of numbers in which each term after the first is obtained by adding a common difference to the preceding term.
The general form of the nth term of an arithmetic sequence is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{General form of the $n$th term of an arithmetic sequence}}\\\\a_n=a+(n-1)d\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a_n$ is the nth term.}\\ \phantom{ww}\bullet\;\textsf{$a$ is the first term.}\\\phantom{ww}\bullet\;\textsf{$d$ is the common difference between terms.}\\\phantom{ww}\bullet\;\textsf{$n$ is the position of the term.}\\\end{array}}[/tex]
[tex]\dotfill[/tex]
Given arithmetic sequence:
[tex]33, \;28, \;23, \;18[/tex]
The first term is a = 33.
To find the common difference, subtract any term from the term that follows it. Let's subtract the second term from the third term:
[tex]d=23-28\\\\d=-5[/tex]
To write the nth term equation for the given arithmetic sequence, substitute the first term a = 33 and the common difference d = -5 into the general form:
[tex]a_n=33+(n-1)(-5)[/tex]
Simplify:
[tex]a_n=33-5n+5\\\\a_n=38-5n[/tex]
Therefore, the nth term equation of the given arithmetic sequence is:
[tex]\Large\boxed{\boxed{a_n=38-5n}}[/tex]
To find the 18th term, substitute n = 18 into the nth term equation:
[tex]a_{18}=38-5(18)\\\\a_{18}=38-90\\\\a_{18}=-52[/tex]
Therefore, the 18th term of the given arithmetic sequence is:
[tex]\Large\boxed{\boxed{a_{18}=-52}}[/tex]
[tex]\dotfill[/tex]
Given arithmetic sequence:
[tex]\dfrac{1}{12},\;\dfrac{1}{3}, \;\dfrac{7}{12},\; \dfrac{5}{6}[/tex]
The first term is a = 1/12.
To find the common difference, subtract any term from the term that follows it. Let's subtract the first term from the second term:
[tex]d=\dfrac{1}{3}-\dfrac{1}{12} \\\\\\ d=\dfrac{1\cdot 4}{3\cdot 4}-\dfrac{1}{12} \\\\\\ d=\dfrac{4}{12}-\dfrac{1}{12} \\\\\\ d=\dfrac{4-1}{12} \\\\\\ d=\dfrac{3}{12} \\\\\\ d=\dfrac{3\div 3}{12\div 3} \\\\\\ d=\dfrac{1}{4}[/tex]
To write the nth term equation for the given arithmetic sequence, substitute the first term a = 1/12 and the common difference d = 1/4 into the general form:
[tex]a_n=\dfrac{1}{12}+(n-1)\dfrac{1}{4}[/tex]
Simplify:
[tex]a_n=\dfrac{1}{12}+\dfrac{1}{4}n-\dfrac{1}{4}\\\\\\a_n=\dfrac{1}{4}n+\dfrac{1}{12}-\dfrac{1\cdot 3}{4\cdot 3}\\\\\\ a_n=\dfrac{1}{4}n+\dfrac{1}{12}-\dfrac{3}{12}\\\\\\ a_n=\dfrac{1}{4}n+\dfrac{1-3}{12}\\\\\\ a_n=\dfrac{1}{4}n-\dfrac{2}{12}\\\\\\ a_n=\dfrac{1}{4}n-\dfrac{1}{6}[/tex]
Therefore, the nth term equation of the given arithmetic sequence is:
[tex]\Large\boxed{\boxed{a_n=\dfrac{1}{4}n-\dfrac{1}{6}}}[/tex]
To find the 50th term, substitute n = 50 into the nth term equation:
[tex]a_{50}=\dfrac{1}{4}(50)-\dfrac{1}{6}\\\\\\ a_{50}=\dfrac{25}{2}-\dfrac{1}{6}\\\\\\ a_{50}=\dfrac{25\cdot 3}{2\cdot 3}-\dfrac{1}{6}\\\\\\ a_{50}=\dfrac{75}{6}-\dfrac{1}{6}\\\\\\ a_{50}=\dfrac{75-1}{6}\\\\\\ a_{50}=\dfrac{74}{6}\\\\\\ a_{50}=\dfrac{37}{3}[/tex]
Therefore, the th term of the given arithmetic sequence is:
[tex]\Large\boxed{\boxed{a_{50}=\dfrac{37}{3}}}[/tex]
[tex]\dotfill[/tex]
Given arithmetic sequence:
[tex]-2023, \; -2020,\;-2017[/tex]
The first term is a = -2023.
To find the common difference, subtract any term from the term that follows it. Let's subtract the first term from the second term:
[tex]d=-2020-(-2023)\\\\d=-2020+2023\\\\d=3[/tex]
To write the nth term equation for the given arithmetic sequence, substitute the first term a = -2023 and the common difference d = 3 into the general form:
[tex]a_n=-2023+(n-1)3[/tex]
Simplify:
[tex]a_n=-2023+3n-3\\\\a_n=3n-2026[/tex]
Therefore, the nth term equation of the given arithmetic sequence is:
[tex]\Large\boxed{\boxed{a_n=3n-2026}}[/tex]
To find the 2023rd term, substitute n = 2023 into the nth term equation:
[tex]a_{2023}=3(2023)-2026\\\\a_{2023}=6069-2026\\\\a_{2023}=4043[/tex]
Therefore, the 2023rd term of the given arithmetic sequence is:
[tex]\Large\boxed{\boxed{a_{2023}=4043}}[/tex]