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Answer :
To solve these questions, we need to understand each part individually involving arithmetic progressions (AP).
Find the 6th term from the end of the AP 17, 14, 11, ..., (-40):
An arithmetic progression (AP) is a sequence of numbers in which the difference of any two successive members is a constant. Here, the first term (a) is 17, and the common difference (d) is -3 (since 14 - 17 = -3).
The last term of the sequence is -40.
First, find the total number of terms (n) in the sequence:
[tex]a_n = a + (n - 1)d \rightarrow -40 = 17 + (n - 1)(-3)[/tex]
Solving for n:
[tex]-40 = 17 - 3n + 3 \rightarrow -60 = -3n \rightarrow n = 20[/tex]
So, there are 20 terms in this sequence.
The 6th term from the end is the same as the (20 - 6 + 1) = 15th term from the start.
Find the 15th term:
[tex]a_{15} = a + 14d = 17 + 14(-3) = 17 - 42 = -25[/tex]
Therefore, the 6th term from the end is -25.
Is 184 a term of the AP 3, 7, 11, 15, ...?
The first term (a) here is 3, and the common difference (d) is 4 (since 7 - 3 = 4).
To check if 184 is a term, use the general term formula of an AP:
[tex]a_n = a + (n - 1)d \rightarrow 184 = 3 + (n - 1)(4)[/tex]
Solving for n:
$184 = 3 + 4n - 4 \rightarrow 184 = 4n - 1 \rightarrow 185 = 4n \rightarrow n = 46.25$
Since n is not an integer, 184 is not a term of the AP.
Is -150 a term of the AP 11, 8, 5, 2, ...?
The first term (a) is 11, and the common difference (d) is -3 (since 8 - 11 = -3).
To check if -150 is a term:
[tex]a_n = a + (n - 1)d \rightarrow -150 = 11 + (n - 1)(-3)[/tex]
Solving for n:
[tex]-150 = 11 - 3n + 3 \rightarrow -164 = -3n \rightarrow n = 54.6667[/tex]
Since n is not an integer, -150 is not a term of the AP.
Which term of the AP 121, 117, 113, ... is its first negative term?
Here, the first term (a) is 121, and the common difference (d) is -4 (since 117 - 121 = -4).
We want to find the first term (n) that is negative:
[tex]a_n = a + (n - 1)d < 0 \rightarrow 121 + (n - 1)(-4) < 0[/tex]
Solving for n:
$121 - 4n + 4 < 0 \rightarrow 125 < 4n \rightarrow n > 31.25$
Therefore, the next integer is n = 32.
Thus, the 32nd term is the first negative term of this AP.
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