Respuesta :

koredeyusuf799

Step-by-step explanation:

since it is geometric sequence we will use the formula

[tex]tn = {a \times r}^{n - 1} [/tex]

a = 2

[tex]r = - \frac{1}{4} [/tex]

The first term

T1(a) = 2

The second Term

[tex]t2 = {a \times r}^{2 - 1} = {a \times r}^{1} [/tex]

[tex]t2 = {2 \times - \frac{1}{4} }^{1} = - \frac{1}{2} [/tex]

The third term

[tex]t3 = {a \times r}^{3 - 1} = {a \times r}^{2} [/tex]

[tex]t3 = {2 \times - \frac{1}{4} }^{2} = 2 \times - \frac{1}{16} = \frac{1}{8} [/tex]

The fourth term

[tex]t4 = {a \times r}^{4 - 1} = {a \times r}^{3} [/tex]

[tex]t4 = {2 \times - \frac{1}{4} }^{3} = 2 \times - \frac{1}{64} = - \frac{1}{32} [/tex]

The fifth term

[tex]t5 = {a \times r}^{5 - 1} = {a \times r}^{4} [/tex]

[tex]t5 = {2 \times - \frac{1}{4} }^{4} = 2 \times - \frac{1}{256} = - \frac{1}{128} [/tex]

i hope all these helped

semsee45

Answer:

[tex]2,\; -\dfrac{1}{2},\; \dfrac{1}{8},\; -\dfrac{1}{32},\; \dfrac{1}{128}[/tex]

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Geometric sequence}\\\\$a_n=ar^{n-1}$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\\phantom{ww}$\bullet$ $r$ is the common ratio.\\\phantom{ww}$\bullet$ $a_n$ is the $n$th term.\\\phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}[/tex]

Given:

  • [tex]a=2[/tex]
  • [tex]r=-\dfrac{1}{4}[/tex]

Substitute the given values of a and r into the formula to create an equation for the nth term:

[tex]a_n=2\left(-\dfrac{1}{4}\right)^{n-1}[/tex]

To find the first 5 terms of the geometric sequence, substitute n = 1 through 5 into the equation.

[tex]\begin{aligned}\implies a_1 & =2\left(-\dfrac{1}{4}\right)^{1-1}\\& =2\left(-\dfrac{1}{4}\right)^{0}\\& =2\left(1\right)\\&=2\end{aligned}[/tex]

[tex]\begin{aligned}\implies a_2 & =2\left(-\dfrac{1}{4}\right)^{2-1}\\& =2\left(-\dfrac{1}{4}\right)^{1}\\& =2\left(-\dfrac{1}{4}\right)\\&=-\dfrac{1}{2}\end{aligned}[/tex]

[tex]\begin{aligned}\implies a_3 & =2\left(-\dfrac{1}{4}\right)^{3-1}\\& =2\left(-\dfrac{1}{4}\right)^{2}\\& =2\left(\dfrac{1}{16}\right)\\&=\dfrac{1}{8}\end{aligned}[/tex]

[tex]\begin{aligned}\implies a_4 & =2\left(-\dfrac{1}{4}\right)^{4-1}\\& =2\left(-\dfrac{1}{4}\right)^{3}\\& =2\left(-\dfrac{1}{64}\right)\\& =-\dfrac{1}{32}\end{aligned}[/tex]

[tex]\begin{aligned}\implies a_5 & =2\left(-\dfrac{1}{4}\right)^{5-1}\\& =2\left(-\dfrac{1}{4}\right)^{4}\\& =2\left(\dfrac{1}{256}\right)\\& =\dfrac{1}{128}\end{aligned}[/tex]

Therefore, the first 5 terms of the given geometric sequence are:

[tex]2,\; -\dfrac{1}{2},\; \dfrac{1}{8},\; -\dfrac{1}{32},\; \dfrac{1}{128}[/tex]

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