Question

Find an explicit rule for the nth term of the sequence.

Answer 1

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 = 7

Hence, 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 = -7

Hence, 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 = -7

Hence, the given formula DO NOT satisfy the given sequence.

So, the ONLY explicit rule for the nth term of the sequence is $a_n = 7 \times (-1)^{n-1}$.