The sum of the first four terms of the sequence is 22.
In this question,
The formula of sum of linear sequence is
[tex]S_n =\frac{n}{2}(2a+(n-1)d)[/tex]
The sum of the first ten terms of a linear sequence is 145
⇒ [tex]S_{10} =\frac{10}{2}(2a+(10-1)d)[/tex]
⇒ 145 = 5 (2a+9d)
⇒ [tex]\frac{145}{5} =2a+9d[/tex]
⇒ 29 = 2a + 9d ------- (1)
The sum of the next ten term is 445, so the sum of first twenty terms is
⇒ 145 + 445
⇒ [tex]S_{20} =\frac{20}{2}(2a+(20-1)d)[/tex]
⇒ 590 = 10 (2a + 19d)
⇒ [tex]\frac{590}{10}=2a+19d[/tex]
⇒ 59 = 2a + 19d -------- (2)
Now subtract (2) from (1),
⇒ 30 = 10d
⇒ d = [tex]\frac{30}{10}[/tex]
⇒ d = 3
Substitute d in (1), we get
⇒ 29 = 2a + 9(3)
⇒ 29 = 2a + 27
⇒ 29 - 27 = 2a
⇒ 2 = 2a
⇒ a = [tex]\frac{2}{2}[/tex]
⇒ a = 1
Thus, sum of first four terms is
⇒ [tex]S_4 =\frac{4}{2}(2(1)+(4-1)(3))[/tex]
⇒ [tex]S_4 =2(2+(3)(3))[/tex]
⇒ S₄ = 2(2+9)
⇒ S₄ = 2(11)
⇒ S₄ = 22.
Hence we can conclude that the sum of the first four terms of the sequence is 22.
Learn more about sum of sequence of n terms here
https://brainly.com/question/20385181
#SPJ4