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  1. What are the first five terms of the sequence given by the formula an = 5n + 1?

1, 5, 9, 13, 17

6, 11, 16, 21, 26

5, 10, 15, 20, 25

5, 6, 7, 8, 9

  1. Write a recursive definition for the sequence 8, 6, 4, 2, …

a1 = 8; an= an–1 − 2

a1 = 2; an= an–1 + 6

a1 = 8; an= an–1 + 2

a1 = 2; an= an–1 + 8

  1. Write an explicit definition for the sequence −2, 1, 4, 7, …

an = –2 + 3n

an = –2 + (n – 3)

an = –2 + (n – 3)

an = –2 + 3 (n – 1)

Verified Answer :

  1. a n = 5 n  + 1
    First 5 terms are:
    B ) 6, 11, 16, 21, 26.
    2.  The sequence is :  8, 6, 4, 2, ...
    A ) a 1 = 8;  a n = a n-1 - 2
    3.  The sequence:  - 2, 1, 4, 7,...
    a 1 = - 2,  common difference: d = 3
    D ) a n = - 2 + 3 ( n - 1 )

The correct answers are:

(1) 6, 11, 16, 21, 26 (Option B)

(2)\(~a_1 = 8\); \(~a_n = a_{n-1} - 2~\) (Option A)

(3)\(~a_n = -2 + 3(n-1)~~~~\) (Option D)

Explanations:

(1) Given Sequence:

\(a_n = 5n + 1\)

Now in order to find the first 5 terms, we need to put n=1,2,3,4,5 in the above sequence and solve.

For n=1: \(a_1 = 5(1) + 1 = 6\)

For n=2: \(a_2 = 5(2) + 1 = 11\)

For n=3: \(a_3 = 5(3) + 1 = 16\)

For n=4: \(a_4 = 5(4) + 1 = 21\)

For n=5: \(a_5 = 5(5) + 1 = 26\)

Hence, the first five terms are: 6, 11, 16, 21, 26 (Option B)

(2) Given Sequence:

8, 6, 4, 2, …

Now to find the recursive definition, we need to adopt trial-and-error approach.

As, \(a_1 = 8\) (meaning the first element of the sequence is 8), the second or nth value of the sequence can be found by using the following formula:

\(a_n = a_{n-1} - d\) --- (1)

Where, n = the index of the number in a sequence

d = difference between two consecutive numbers = 8-6 = 2

Now,

The second number of the sequence has to be 6 by using (1). Put n = 2 and d = 2 in (1):

\(a_2 = a_{2-1} - 2\)

\(a_2 = a_{1} - 2\)

Since \(a_1 = 8\), therefore,

\(a_2 = 8 - 2 = 6\) (correct)

Hence the correct answer is \(a_1 = 8\); \(~a_n = a_{n-1} - 2~\) (Option A)

(3) Given Sequence:

−2, 1, 4, 7, …

To find the explicit definition, use the following formula:

\(a_n = a_1 + (n-1)*d\) --- (X)

Where,

\(a_n = nth~term~of~the~sequence \\a_1 = 1st~term~of~the~sequence = -2 \\d = common~difference = 4-1 = 7-4 = 3 \ = index~of~a~number~in~a~sequence \\\)

Plug in the values in (X):

(X)=> \(a_n = -2 + (n-1)*3~~~~\) (Option D)