Respuesta :

koredeyusuf799

Step-by-step explanation:

since it is geometric sequence we will use the formula

[tex]tn = {ar}^{n - 1} [/tex]

a = 6

r = 4

The first term

T1(a) = 6

The second term

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

[tex]t2 = {6 \times 4}^{1 } = 24[/tex]

The Third Term

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

[tex]t3 = {6 \times 4}^{2} = 6 \times 16 = 96[/tex]

The fourth term

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

[tex]t4 = {6 \times 4}^{3} = 6 \times 64 = 384[/tex]

The fifth term

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

[tex]t5 = {6 \times 4}^{4} = 6 \times 256 = 1,536[/tex]

i hope these helped

semsee45

Answer:

6, 24, 96, 384, 1536, ...

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:

  • a = 6
  • r = 4

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

[tex]a_n=6(4)^{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&=6(4)^{1-1}\\&=6(4)^{0}\\&=6(1)\\&=6\end{aligned}[/tex]

[tex]\begin{aligned}\implies a_2&=6(4)^{2-1}\\&=6(4)^{1}\\&=6(4)\\&=24\end{aligned}[/tex]

[tex]\begin{aligned}\implies a_3&=6(4)^{3-1}\\&=6(4)^{2}\\&=6(16)\\&=96\end{aligned}[/tex]

[tex]\begin{aligned}\implies a_4&=6(4)^{4-1}\\&=6(4)^{3}\\&=6(64)\\&=384\end{aligned}[/tex]

[tex]\begin{aligned}\implies a_5&=6(4)^{5-1}\\&=6(4)^{4}\\&=6(256)\\&=1536\end{aligned}[/tex]

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

  • 6, 24, 96, 384, 1536, ...

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