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3 years ago

Find the 4th term in the sequence with the following definition: a1=-2 an an=2an-1-3

Mathematics

1 answer:

Alexus [3.1K]3 years ago

7 0

Answer:

The 4th term is -37.

Step-by-step explanation:

The first term of the series is

$a_1=-2$ ,

the second term of the series is

$a_2 =2a_1-3$

$a_2=2(-2)-3\\\\a_2=-7,$

the third term is

$a_3 = 2a_2-3$

$a_3=2(-7)-3$

$a_3=-17$ ,

and the fourth term of the series is

$a_4=2a_3-3\\$

$a_4=2(-17)-3$

$\boxed{a_4=-37}$

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olasank [31]

Answer:

12 bikes

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Step-by-step explanation:

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5 0

3 years ago

2 If A= 4 3 find A1 using elementary row operations.

MA_775_DIABLO [31]

Answer: $A^{-1}=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]$

Step-by-step explanation:

$\left[\begin{array}{cc}2&1\\4&3\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

$\dfrac{1}{2}Row\ 1\rightarrow\left[\begin{array}{cc}1&\frac{1}{2}\\4&3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\0&1\end{array}\right]$

$Row\ 2 -4 \ Row\ 1\rightarrow \left[\begin{array}{cc}1&\frac{1}{2}\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\-2&1\end{array}\right]$

$Row\ 1-\dfrac{1}{2}\ Row\ 2 \rightarrow \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]$