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3 years ago

How many terms are in the arithmetic sequence 7, 1, −5, …, −161?

Mathematics

1 answer:

uranmaximum [27]3 years ago

5 0

Answer:

Step-by-step explanation:

The n th term of an arithmetic sequence is

• $a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

d = 1 - 7 = - 5 - 1 = - 6 and a₁ = 7, $a_{n}$ = - 161, hence

7 - 6(n - 1) = - 161 ← solve for n

7 - 6n + 6 = - 161

13 - 6n = - 161 ( subtract 13 from both sides )

- 6n = - 174 ( divide both sides by - 6 )

n = 29

There are 29 terms in the sequence

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The polynomial x^2+3x-1 is a factor of x^4+3x^3-2x^2-3x+1 true or false?

pashok25 [27]

Answer:

Yes, it is true that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ .

Step-by-step explanation:

Let us try to factorize $x^4+3x^3-2x^2-3x+1$

$x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3$

Let us try to make a whole square of the given terms:

$\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\$

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Formula used above:

$a^{2} -2 \times a \times b +b^{2} = (a-b)^2$

In the above equation, we had $a = x, b = 1$ .

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Further solving the above equation, taking $3x$ common out of $3x^3-3x$

$\Rightarrow (x^2-1)^2+3x(x^2-1)\\$

Taking $(x^{2} -1)$ common out of the above term:

$\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)$

So, the two factors are $(x^2-1)\ and\ (x^2+3x-1)$ .

$\therefore$ The statement that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ is True.

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