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Fofino [41]
3 years ago
5

If the second term of an arithmetic sequence is 5 and the fourth term is 12, what is the 37th term?

Mathematics
2 answers:
neonofarm [45]3 years ago
7 0
By using the information in the question, find a equation in the y=mx+b form. Then, plug in 37 as the x value and solve for y.
Setler [38]3 years ago
5 0
Aritmetic sequence
an=a1+d(n-1)
a1=first term
d=common differnce
an=nth term


given
a2=5
a4=12

so
a2=a1+d(2-1)=5
a4=a1+d(4-1)=12

a1+d=5
a1+3d=12
so 12-5=(a1+d)-(a1+3d)=2d=7
2d=7
divide by 2
d=3.5

a2=a1+3.5(1)
5=a1+3.5
minus 3.5 both sides
1.5=a1

an=1.5+3.5(n-1)
37th term
a37=1.5+3.5(37-1)
a37=1.5+3.5(36)
a37=1.5+126
a37=127.5


37th term is 127.5
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Step-by-step explanation:

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8 0
1 year ago
1.94x - 10/8x + 4.32 = [2(1.15x+8.5)]
kipiarov [429]
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3 years ago
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Thepotemich [5.8K]

Answer:

the answer would be 135 meters

Step-by-step explanation:

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A line passes through the points (2,-1) and (1,-5).

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The equation of the line.

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y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)

The line passes through the points (2,-1) and (1,-5). So, the equation of the line is

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