Enter a recursive rule for the geometric sequence. 6, −18, 54, −162, ...

Mathematics

1 answer:

svlad2 [7]3 years ago

5 0

Answer:

The required recursive formula is:

$a_n=(-3)\times a_{n-1}$

Step-by-step explanation:

We are given a geometric sequence as:

6,-18,54,-162,.....

Clearly after looking at different terms of the sequence we could observe that the sequence is an geometric progression (G.P.) with common ratio= -3 denoted by r.

Let $a_n$ represents the nth term of the sequence.

This means that:

$a_1=6, a_2=-18, a_3=54, a_4=-162,......$

As the common ratio is -3.

so,

$a_1=6\\\\a_2=-18=(-3)\times a_1\\\\a_3=54=(-3)\times a_2\\\\.\\.\\.\\.\\.\\.\\.\\.\\.a_n=(-3)\times a_{n-1}$

Hence, the required recursive formula for the geometric sequence is:

$a_n=(-3)\times a_{n-1}$

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Solve the following quadratic equation by the square root method. (Y-9)^2=64

enot [183]

Hello!

To solve this, first perform the opposite operation for the last operation (on the left side) on both sides. The last operation of the left side is squaring. Therefore, square root both sides.

$(y - 9)^{2} = 64$

$y - 9 = ±\sqrt{64}$

$y - 9 = ±8$

Please note that you must include ±. This is because the square root of 64 can be either positive or negative, as a square of either a positive or negative number is positive.

Now, add 9 to both sides.

$y = 9±8$