In a geometric sequence {an}, if a1 = 4 and r = 0.2, the first 4 terms in the sequence are A) {4, .8, .16, .032,...}. Eliminate B) {4, 20, 100, 500,...}. C) {4, 4.2, 4.4, 4.6,...}. D) {4, 3.8, 3.6, 3.4,...}.

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Hey man, your answer is (A) I took the test!!!



The first 4 terms in the sequence are :

A) { 4 , 0.8 , 0.16 , 0.032 , . . . }

Further explanation

Firstly , let us learn about types of sequence in mathematics.

Arithmetic Progression is a sequence of numbers in which each of adjacent numbers have a constant difference.

[tex]\boxed{T_n = a + (n-1)d}[/tex]

[tex]\boxed{S_n = \frac{1}{2}n ( 2a + (n-1)d )}[/tex]

Tn = n-th term of the sequence

Sn = sum of the first n numbers of the sequence

a = the initial term of the sequence

d = common difference between adjacent numbers

Geometric Progression is a sequence of numbers in which each of adjacent numbers have a constant ration.

[tex]\boxed{T_n = a ~ r^{n-1}}[/tex]

[tex]\boxed{S_n = \frac{a( 1 - r^n ) }{1 - r}}[/tex]

Tn = n-th term of the sequence

Sn = sum of the first n numbers of the sequence

a = the initial term of the sequence

r = common ratio between adjacent numbers

Let us now tackle the problem!

Given:

a = 4

r = 0.2

Solution:

[tex]T_n = a ~ r^{n-1}[/tex]

[tex]T_1 = 4 \times 0.2^{1-1} = 4 \times 1 = 4[/tex]

[tex]T_2 = 4 \times 0.2^{2-1} = 4 \times 0.2 = 0.8[/tex]

[tex]T_3 = 4 \times 0.2^{3-1} = 4 \times 0.04 = 0.16[/tex]

[tex]T_4 = 4 \times 0.2^{4-1} = 4 \times 0.008 = 0.032[/tex]

Therefore , the first 4 terms in the sequence are :

A) { 4 , 0.8 , 0.16 , 0.032 , . . . }

Learn more

  • Geometric Series : https://brainly.com/question/4520950
  • Arithmetic Progression : https://brainly.com/question/2966265
  • Geometric Sequence : https://brainly.com/question/2166405

Answer details

Grade: Middle School

Subject: Mathematics

Chapter: Arithmetic and Geometric Series

Keywords: Arithmetic , Geometric , Series , Sequence , Difference , Term

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