The the sixth term of the arithmetic sequence whose 31st and 73rd terms are 18 and 46, respectively is 4/3.
According to the given question.
The 31st term of the arithmetic sequence is 18
And the 73rd term of the arithmetic sequence is 46
As, we know that the general term of the arithmetic sequence is given by
[tex]a_{n}[/tex] = a + (n-1)
Where,
[tex]a_{n}[/tex] is the nth term
a is the first term and
d is the common difference
So, the 31 term is given by
[tex]a_{31}[/tex] = a + (31 - 1)d
⇒ 18 = a + 30d ...(i)
Similarly, the 73rd term is given by
46 = a + (73 - 1)d
⇒ 46 = a + 72d..(ii)
Subtract equation (i) from (ii)
46 - 18 = 72d - 30d
⇒ 28 = 42d
⇒ d = 28/42
⇒ d = 4/6
⇒ d = 2/3
Now,
18 = a + 30(2/3)
⇒ 18 = a + 20
⇒ a = -2
Therefore, the sixth term of the arithmetic sequence whose 31st and 73rd terms are 18 and 46, respectively is given by
[tex]a_{6}[/tex] = -2 + (6 -1)(2/3)
⇒ [tex]a_{6}[/tex] = -2 +10/3
⇒ [tex]a_{6}[/tex] = 4/3
Hence, the the sixth term of the arithmetic sequence whose 31st and 73rd terms are 18 and 46, respectively is 4/3.
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