Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
Given series is
[tex]\rm :\longmapsto\:65, \: 62, \: 59, \: - - - - [/tex]
Since, its an AP series with
First term, a = 65
Common difference, d = - 3
Let assume that nth term of the series be first negative term.
Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,
↝ nᵗʰ term of an arithmetic sequence is,
[tex]\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{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,
[tex]\red{\rm :\longmapsto\:a_n < 0}[/tex]
[tex]\rm :\longmapsto\:a + (n - 1)d < 0[/tex]
[tex]\rm :\longmapsto\:65+ (n - 1)( - 3) < 0[/tex]
[tex]\rm :\longmapsto\:65 - 3n + 3 < 0[/tex]
[tex]\rm :\longmapsto\:68 - 3n < 0[/tex]
[tex]\rm :\longmapsto\:- 3n < - 68[/tex]
[tex]\rm \implies\:n > \dfrac{68}{3} [/tex]
[tex]\rm \implies\:n > 22.66[/tex]
[tex]\bf\implies \: {23}^{rd} \: term \: is \: first \: negative \: term.[/tex]
So, Option (b) is correct.
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↝ Sum of n terms of an arithmetic sequence is,
[tex]\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}[/tex]
Wʜᴇʀᴇ,
Sₙ is the sum of n terms of AP.
a is the first term of the sequence.
n is the no. of terms.
d is the common difference.
