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A sequence is shown in the graph.

points at 1 comma 48, 2 comma 12, 3 comma 3, and 4 comma 75 hundredths

Assuming the pattern continues, what is the formula for the nth term?

an = 48(4n)
an = 48(0.25n + 1)
an = 4(48n − 1)
an = 48(0.25n − 1)

A Sequence Is Shown In The Graph Points At 1 Comma 48 2 Comma 12 3 Comma 3 And 4 Comma 75 Hundredths Assuming The Pattern Continues What Is The Formula For The class=

Sagot :

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Answer:

  (d)  48(0.25^(n-1))

Step-by-step explanation:

The points shown on the graph have y-values that decrease by a factor of 4 as x-values increase by 1. The first couple are (1, 48), (2, 12). You want the formula for the n-th term.

Geometric sequence

The terms of a geometric sequence have a common ratio (r). If the first term is a1, the general term is ...

  an = a1(r^(n-1))

Application

The given sequence has first term a1 = 48. The common ratio is ...

  r = 12/48 = 1/4 = 0.25

Using these value in the formula for the general term, we find the n-th term to be ...

  an = 48(0.25^(n-1)) . . . . n-th term of the pattern

[tex]\begin{array}{ccccccccc} 1&&2&&3&&4&&5\\\cline{1-9} \\48&,&12&,&3&,&\underset{0.75}{\frac{3}{4}}&&\underset{0.1875}{\frac{3}{16}}\\[2em]\cline{1-9} &&48\left( \frac{1}{4} \right)&&12\left( \frac{1}{4} \right)&&3\left( \frac{1}{4} \right)&&\frac{3}{4} \end{array}~\hfill \begin{array}{llll} a_1=48\\\\ r=\frac{1}{4} \end{array} \\\\[-0.35em] ~\dotfill[/tex]

[tex]n^{th}\textit{ term of a geometric sequence} \\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\stackrel{\textit{first term}}{48}\\ r=\stackrel{\textit{common ratio}}{\frac{1}{4}\to 0.25} \end{cases}\implies a_n=48\left( 0.25 \right)^{n-1}[/tex]

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