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3 years ago

The first term of an arithmetic sequence is 4 and the fifth term is 16. What is the nth term of the sequence?

1 answer:

8 0

First term, a = 4

Fifth term = 16

The nth term of arithmetic sequence : a + (n - 1)*d

For fifth term, n = 5

a + (n - 1)*d

Substitute n = 5, a = 4, d = ?, 5th term = 16

4 + (5 - 1)d = 16

4 + 4d = 16

4d = 16 - 4

4d = 12 Divide by 4

d = 12/4

d = 3.

Recall the nth term is: a + (n - 1)*d = 4 + (n - 1)*3

4 + (n - 1)*3

4 + 3n - 3

3n + 4 - 3

3n + 1

So the nth term is 3n + 1. Option B.

Hope this helps

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Penelope determined the solutions of the quadratic function by completing the square. F(x) = 4x2 8x 1 –1 = 4x2 8x –1 = 4(x2 2x)

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By solving the quadratic equation using completing the square method we got that Penelope should have added 4 to both sides instead of adding 1.

<h3>What is quadratic equation ?</h3>

Any equation of the form $ax^2+bx+c=0$ where x is variable and a, b, and c are any real numbers where a ≠ 0 is called quadratic equation .

Given work of Penelope is

$\begin{aligned}&f(x)=4 x^{2}+8 x+1 \\&-1=4 x^{2}+8 x \\&-1=4\left(x^{2}+2 x\right) \\&-1+1=4\left(x^{2}+2 x+1\right) \\&0=4(x+2)^{2} \\&0=(x+2)^{2} \\&0=x+2 \\&-2=x\end{aligned}$

Now we can see that Penelope determined the solutions of the quadratic function by completing the square. so he must have been done as following

$\begin{aligned}&f(x)=4 x^{2}+8 x+1 \\&-1=4 x^{2}+8 x \\&-1=4\left(x^{2}+2 x\right) \\&-1+4=4\left(x^{2}+2 x+1\right) \\&3=4(x+2)^{2} \\&\frac{3}{4} =x+2)^{2} \\&\frac{\pm\sqrt3}{2} =x+2 \\&\frac{-2\pm\sqrt3}{2} =x\end{aligned}$

By solving the quadratic equation using completing the square method we got that Penelope should have added 4 to both sides instead of adding 1.

To learn more about quadratic equations visit : brainly.com/question/1214333

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