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3 years ago

Indicate the general rule for the arithmetic sequence with a3 = -12 and a8 = -37.

Mathematics

1 answer:

jeka57 [31]3 years ago

5 0

Answer:

Option A is correct

General rule for arithmetic sequence with $a_3 = -12$ and $a_8 = -37$ is; $a_n=-2+(n-1)(-5)$

Step-by-step explanation:

Arithmetic sequence states that a sequence where the difference between each successive pair of terms is the same.

The general rule for the arithmetic sequence is given by;

$a_n=a+(n-1)d$ where

a represents the first term

d represents the common difference and

n represents the number of terms.

Given: $a_3 = -12$ and $a_8 = -37$

$a_3 = -12$

a+(3-1)d = -12 [Using arithmetic sequence rule]

a + 2d = -12

or we can write this as;

a = -12 - 2d ......[1]

Similarly, for $a_8 = -37$ we have;

$a+(8-1)d = -37$

a+7d = -37 ......[2]

Substitute equation [1] into [2] to solve for d;

-12 - 2d +7d = -37

Combine like terms;

-12 + 5d = -37

Add both sides 12 we get;

-12 + 5d + 12 = -37 + 12

Simplify:

5d = -25

Divide both sides by 5 we get;

d = -5

Substitute the value of d in equation [1] to solve for a;

a = -2(-5) - 12

a = 10 -12 = -2

∴ a = -2

therefore, the general rule for the arithmetic sequence with $a_3 = -12$ and $a_8 = -37$ is, $a_n=-2+(n-1)(-5)$

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Read 2 more answers

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

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$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

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