Show all work. find the missing terms of the arithmetic sequence... , 4, ?, ?, ?, 120, ...

Mathematics

1 answer:

denis23 [38]3 years ago

6 0

Answer:

-25, 4, 33, 62, 91, 120, ...

Step-by-step explanation:

<h3>Given</h3>

<u>The arithmetic sequence</u>

.. , 4, ?, ?, ?, 120, ...

Missing terms

<h3>Solution</h3>

<u>Let the first term a₁ = x, and common difference = d then we have:</u>

a₁ = x
a₂ = x + d = 4
...
a₆ = x + 5d = 120

<u>The difference of the 6th and 2nd terms:</u>

a₆ - a₂ =
a + 5d - x - d = 120 - 4
4d = 116
d = 116/4
d = 29

<u>Then we get the first term and other missing terms:</u>

x = 4 - 29 = -25
a₁ = -25

-25, 4, 4+29 = 33, 33+29 = 62, 62 + 29= 91, 91+29 = 120, ...

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Use synthetic division to find P (-10) for P(x)=2x^3+14x^2-58x

Savatey [412]

The polynomial remainder theorem states that the remainder upon dividing a polynomial $p(x)$ by $x-c$ is the same as the value of $p(c)$ , so to find $p(-10)$ you need to find the remainder upon dividing

$\dfrac{2x^3+14x^2-58x}{x+10}$

You have

..... | 2 ... 14 ... -58
-10 | ... -20 ... 60
--------------------------
..... | 2 ... -6 .... 2

So the quotient and remainder upon dividing is

$\dfrac{2x^3+14x^2-58x}{x+10}=2x-6+\dfrac2{x+10}$

with a remainder of 2, which means $p(-10)=2$ .