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2 years ago

PLEASE HELP!!!!! 3, 8, 13, 18, 23, .... The recursive formula for this sequence is:

Mathematics

1 answer:

rodikova [14]2 years ago

8 0

Answer:

a₈ = 37

Step-by-step explanation:

The given arithmetic sequence is: 3, 8, 13, 18, 23, . . .

The recursive formula for the sequence is: $a_n = a_{n - 1} + 5$

Here, $a_n$ represents the $n^{th}$ of the sequence.

And, $a_{n - 1}$ represents the $(n - 1)^{th}$ of the sequence.

'+5' denotes that '5' is added to the $(n - 1)^{th}$ term to get the $n^{th}$ term. In other words, the difference between two consecutive numbers in the sequence is 5.

Now, we are asked to find a₈ i.e., n =8.

Substituting in the recursive formula we get: a₈ = a₍₈₋ ₁₎ + 5 = a₇ + 5.

So, to determine a₈ we need to know a₇. From the sequence we see that a₅ = 23.

⇒ a₆ = 23 + 5 = 28.

⇒ a₇ = 28 + 5 = 32.

⇒ a₈ = 32 + 5 = 37.

Therefore, the $8^{th}$ term of the sequence is 37.

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Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

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