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3 years ago

What is the 9th term of the sequence? 3,-12,48,-192,

Mathematics

2 answers:

Inessa [10]3 years ago

3 0

D, 196,608.

The sequence is multiplying by -4 every time.

chubhunter [2.5K]3 years ago

3 0

Answer: The correct option is (D) 196608.

Step-by-step explanation: We are given to find the 9th term of the following sequence :

3, -12, 48, -192, . . .

Let a(n) denote the n-th term of the given sequence.

Then, a(1) = 3, a(2) = -12, a(3) = 48, a(4) = -192, . . .

We see that

$\dfrac{a(2)}{a(1)}=\dfrac{-12}{3}=-4,\\\\\\\dfrac{a(3)}{a(2)}=\dfrac{48}{-12}=-4,\\\\\\\dfrac{a(4)}{a(3)}=\dfrac{-192}{48}=-4,~~.~~.~~.$

So, we get

$\dfrac{a(2)}{a(1)}=\dfrac{a(3)}{a(2)}=\dfrac{a(4)}{a(3)}=~~.~~.~~.~~=-4.$

That is, the given sequence is a GEOMETRIC one with first term a = 3 and common ratio d= -4.

We know that

the n-th term of an geometric sequence with first term a and common ratio r is given by

$a(n)=ar^{n-1}.$

Therefore, the 9th term of the given sequence is

$a(9)=ar^{9-1}=3\times(-4)^8=3\times 65536=196608.$

Thus, the 9th term of the given sequence is 196608.

Option (D) is CORRECT.

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The zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i and the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

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The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

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So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

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f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

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(5y - 4)(5y + 4)(25y^2 + 16) = 0

Expand the equation

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Solve the factors of the equation

So, we have

y = 4/5, y = -4/5 and y = ±4/5√i

Hence, the zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i

How to write the polynomial as a product of the linear factors?

In (a), we have

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Hence, the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)