Answer:
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[tex]\large\underline{\sf{Solution-}}[/tex]
Given AP series is
[tex]\sf \: 15, \: 12, \: 9, \: ... \\ \\ [/tex]
Here,
First term of an AP, a = 15
Common difference, d = 12 - 15 = - 3
Let assume that nth term of an AP series, [tex]a_n [/tex] = - 90.
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 sequence.
n is the no. of terms.
d is the common difference.
Tʜᴜs, on substituting the values, we get
[tex]\sf \: - 90 = 15 + (n - 1)( - 3) \\ \\ [/tex]
[tex]\sf \: - 90 - 15 = - 3n + 3 \\ \\ [/tex]
[tex]\sf \: - 105= - 3n + 3 \\ \\ [/tex]
[tex]\sf \: 3n = 105 + 3 \\ \\ [/tex]
[tex]\sf \: 3n = 108 \\ \\ [/tex]
[tex]\sf \: \implies \: n = 36 \\ \\ [/tex]
Hence, 36th term of an AP 15, 12, 9... is - 90.
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Answer:
-54 plzz mark me as brainalist
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given AP series is
[tex]\sf \: 15, \: 12, \: 9, \: ... \\ \\ [/tex]
Here,
First term of an AP, a = 15
Common difference, d = 12 - 15 = - 3
Let assume that nth term of an AP series, [tex]a_n [/tex] = - 90.
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 sequence.
n is the no. of terms.
d is the common difference.
Tʜᴜs, on substituting the values, we get
[tex]\sf \: - 90 = 15 + (n - 1)( - 3) \\ \\ [/tex]
[tex]\sf \: - 90 - 15 = - 3n + 3 \\ \\ [/tex]
[tex]\sf \: - 105= - 3n + 3 \\ \\ [/tex]
[tex]\sf \: 3n = 105 + 3 \\ \\ [/tex]
[tex]\sf \: 3n = 108 \\ \\ [/tex]
[tex]\sf \: \implies \: n = 36 \\ \\ [/tex]
Hence, 36th term of an AP 15, 12, 9... is - 90.