Write an explicit formula for an arithmetic sequence whose common difference is -3, and first term is -9.

Mathematics

2 answers:

schepotkina [342]3 years ago

7 0

Answer:

The required formula is $a_n =-3n-6$ .

Step-by-step explanation:

Given : An arithmetic sequence whose common difference is -3, and first term is -9.

To find : Write an explicit formula for an arithmetic sequence ?

Solution :

An explicit formula for an arithmetic sequence is given by, $a_n = a_1 +d(n -1)$

Where, $a_n$ is the nth term of the sequence

$a_1=-9$ is the first term of the sequence

d=-3 is the common difference

n is the number terms

Substitute the value in the formula,

$a_n =(-9)+(-3)(n -1)$

$a_n =-9-3n+3$

$a_n =-3n-6$

Therefore, the required formula is $a_n =-3n-6$ .

kirza4 [7]3 years ago

5 0

Answer:

an = -9 -3(n -1)

Step-by-step explanation:

The explicit formula for an arithmetic sequence with first term a1 and common difference d is ...

an = a1 +d(n -1)

Putting your numbers into the formula gives ....

an = -9 -3(n-1)

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kompoz [17]

Answer:

$\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}$

Step-by-step explanation:

Your expression is

$\sqrt [3] {-54x^{5}y^{6}}$

Here's how I would simplify it.

$\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}$

$\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}$