Q:

# Find an explicit rule for the nth term of the sequence.7, -7, 7, -7, ... (5 points)Select one:a. an = 7 • (-1)n-1b. an = 7 • (-1)nc. an = 7 • 1n-1d. an = 7 • 1n+1

Accepted Solution

A:
Answer:(a) The ONLY explicit rule for the nth term of the sequence is $$a_n = 7 \times (-1)^{n-1}\\$$.Step-by-step explanation:Here, the given sequence is 7, -7, 7, -7, ... The first term = 7, Second term = -7, Third term =-7 and so on..Now check the given sequence for each given formula, we get:(1)  $$a_n = 7 \times (-1)^{n-1}\\$$Now, for n = 1 : $$a_1 = 7 \times (-1)^{1-1}\\$$                                        $$= 7 \times (-1)^0 = 7 \times 1 = 7 \implies a_1 = 7$$Similarly, for, n = 2:  $$a_2 = 7 \times (-1)^{2-1}\\$$                                        $$= 7 \times (-1)^1 = 7 \times (-1) = -7 \implies a_2 = -7$$Hence, the given formula satisfies the given sequence.(2)  $$a_n = 7 \times (-1)^{n}\\$$Now, for n = 1 : $$a_1 = 7 \times (-1)^{1}\\$$                                        $$= 7 \times (-1)^1 = 7 \times (-1) = -7 \implies a_1 = -7$$But, First term = 7Hence, the given formula DO NOT satisfy the given sequence.(3)  $$a_n = 7 \times (1)^{n-1}\\$$Now, for n = 1 : $$a_1 = 7 \times (1)^{1-1}\\$$                                        $$= 7 \times (1)^0 = 7 \times 1 = 7 \implies a_1 = 7$$Similarly, for, n = 2:  $$a_2 = 7 \times (1)^{2-1}\\$$                                        $$= 7 \times (1)^1 = 7 \times (1) = 7 \implies a_2 = 7$$But, Second term = -7Hence, the given formula DO NOT satisfy the given sequence.(4) $$a_n = 7 \times (1)^{n+1}\\$$Now, for n = 1 : $$a_1 = 7 \times (1)^{1+1}\\$$                                        $$= 7 \times (1)^2 = 7 \times 1 = 7 \implies a_1 = 7$$Similarly, for, n = 2:  $$a_2 = 7 \times (1)^{2+1}\\$$                                        $$= 7 \times (1)^3 = 7 \times (1) = 7 \implies a_2 = 7$$But, Second term = -7Hence, the given formula DO NOT satisfy the given sequence.So, the ONLY explicit rule for the nth term of the sequence is $$a_n = 7 \times (-1)^{n-1}\\$$.