Answer:
The 13th term of the sequence is 120.
Step-by-step explanation:
Given,
[tex]a_1=12\\d=21-12=9\\n=13\\a_1_3=?[/tex]
We know that,
[tex]a_n = a_1 + (n-1)d[/tex]
[tex]\Rightarrow a_{13} = 12 + (13-1) \times 9[/tex]
[tex]\Rightarrow a_{13} = 12 + 12 \times 9[/tex]
[tex]\Rightarrow a_{13} = 12 + 108[/tex]
[tex]\Rightarrow a_{13} = 120[/tex]
[tex]\boxed{\sf\:\bf \: {13}^{th} \: term \: of \: an \: arithmetic \: sequence \: is \: 120 \: }[/tex]
Given sequence is: 12, 21, 30, ...
Now, 21 - 12 = 9, 30 - 21 = 9
So, It means 12, 21, 30, ... forms an arithmetic sequence.
So, we have
First term, a = 12
Common difference, d = 9
Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,
↝ nᵗʰ term of an arithmetic sequence is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \end{gathered}[/tex]
Wʜᴇʀᴇ,
aₙ is the nᵗʰ term.
a is the first term of the progression.
n is the no. of terms.
d is the common difference.
So, using this result, we get
[tex]\bf \: a_{13} \\ [/tex]
[tex]\sf \: = \: a + ({13 - 1})d \\ [/tex]
[tex]\sf \: = \: a + 12d \\ [/tex]
[tex]\sf \: = \: 12 + 12 \times 9 \\ [/tex]
[tex]\sf \: = \: 12 + 108 \\ [/tex]
[tex]\sf \: = \:120 \\ [/tex]
Thus,
[tex]\implies\sf \:\bf \: a_{13} = 120\\ [/tex]
Hence,
[tex]\implies\sf \:\bf \: {13}^{th} \: term \: of \: an \: arithmetic \: sequence \: is \: 120\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
↝ Sum of n terms of an arithmetic progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \end{gathered}[/tex]
↝ nᵗʰ term of an arithmetic progression is,
↝ nᵗʰ term of an arithmetic progression from end is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:l\:-\:(n\:-\:1)\:d}}}}}} \end{gathered}[/tex]
aₙ is the nᵗʰ term
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Answers & Comments
Answer:
The 13th term of the sequence is 120.
Step-by-step explanation:
Given,
[tex]a_1=12\\d=21-12=9\\n=13\\a_1_3=?[/tex]
We know that,
[tex]a_n = a_1 + (n-1)d[/tex]
[tex]\Rightarrow a_{13} = 12 + (13-1) \times 9[/tex]
[tex]\Rightarrow a_{13} = 12 + 12 \times 9[/tex]
[tex]\Rightarrow a_{13} = 12 + 108[/tex]
[tex]\Rightarrow a_{13} = 120[/tex]
Answer:
[tex]\boxed{\sf\:\bf \: {13}^{th} \: term \: of \: an \: arithmetic \: sequence \: is \: 120 \: }[/tex]
Step-by-step explanation:
Given sequence is: 12, 21, 30, ...
Now, 21 - 12 = 9, 30 - 21 = 9
So, It means 12, 21, 30, ... forms an arithmetic sequence.
So, we have
First term, a = 12
Common difference, d = 9
Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,
↝ nᵗʰ term of an arithmetic sequence is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \end{gathered}[/tex]
Wʜᴇʀᴇ,
aₙ is the nᵗʰ term.
a is the first term of the progression.
n is the no. of terms.
d is the common difference.
So, using this result, we get
[tex]\bf \: a_{13} \\ [/tex]
[tex]\sf \: = \: a + ({13 - 1})d \\ [/tex]
[tex]\sf \: = \: a + 12d \\ [/tex]
[tex]\sf \: = \: 12 + 12 \times 9 \\ [/tex]
[tex]\sf \: = \: 12 + 108 \\ [/tex]
[tex]\sf \: = \:120 \\ [/tex]
Thus,
[tex]\implies\sf \:\bf \: a_{13} = 120\\ [/tex]
Hence,
[tex]\implies\sf \:\bf \: {13}^{th} \: term \: of \: an \: arithmetic \: sequence \: is \: 120\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
↝ Sum of n terms of an arithmetic progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \end{gathered}[/tex]
↝ nᵗʰ term of an arithmetic progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \end{gathered}[/tex]
↝ nᵗʰ term of an arithmetic progression from end is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:l\:-\:(n\:-\:1)\:d}}}}}} \end{gathered}[/tex]
Wʜᴇʀᴇ,
aₙ is the nᵗʰ term.
a is the first term of the progression.
n is the no. of terms.
d is the common difference.
aₙ is the nᵗʰ term