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3 years ago

Write the first five terms of the geometric sequence in which a1 = 27 and the common ratio is 4/3

Mathematics

1 answer:

tangare [24]3 years ago

5 0

Answer:

c) 27, 36, 48, 64, 256/3

Step-by-step explanation:

We are given a geometric sequence with;

First term, a = 27
Common ratio, r = 4/3

We are supposed to write the sequence;

We know that a given term in a GP is given by;

$T_n=ar^n^-^1$ , where a is the first term, n is the nth term

Therefore;

$First term, a = 27 \\Second term = ar^1= 27(4/3) = 36\\Third term = ar^2=27(4/3)^2=48\\Fourth term =ar^3=27(4/3)^3=64\\Fifth term = ar^4 = 27(4/3)^4 =\frac{256}{3} \\And so forth$

Therefore;

The geometric sequence is;

$27,36,48,64,\frac{256}{3}$

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3 years ago

What is the value of 2 in 255.6

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3 years ago

Which expressions are equivalent??

Brilliant_brown [7]

Given:

The given expression is $2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)$

We need to determine the equivalent expression.

Option A: -1

Solving the expression, we get;

$\frac{3}{2} x+14-\frac{3}{2} x+15$

Simplifying, we get;

$14+15=29$

Thus, the expression $2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)$ is not equivalent to -1.

Hence, Option A is not the correct answer.

Option B: 29

Solving the expression, we get;

$\frac{3}{2} x+14-\frac{3}{2} x+15$

Simplifying, we get;

$14+15=29$

Thus, the expression $2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)$ is equivalent to 29.

Hence, Option B is the correct answer.

Option C: $2\left(\frac{3}{4} x+7\right)+(-3)\left[\frac{1}{2} x+(-5)\right]$

Let us rewrite the given expression.

$2\left(\frac{3}{4} x+7\right)+(-3)\left[\frac{1}{2} x+(-5)\right]$

Thus, the expression $2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)$ is equivalent to $2\left(\frac{3}{4} x+7\right)+(-3)\left[\frac{1}{2} x+(-5)\right]$

Thus, Option C is the correct answer.

Option D: $2\left(\frac{3}{4} x\right)+2(7)+3\left(\frac{1}{2} x\right)+3(-5)$

Rewriting the expression, we get;

$2\left(\frac{3}{4} x+7\right)+(-3)\left[\frac{1}{2} x+(-5)\right]$

Hence, the expression $2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)$ is equivalent not to $2\left(\frac{3}{4} x\right)+2(7)+3\left(\frac{1}{2} x\right)+3(-5)$

Thus, Option D is not the correct answer.

Option E: $2\left(\frac{3}{4} x\right)+2(7)+(-3)\left(\frac{1}{2} x\right)+(-3)(-5)$

Multiplying the terms within the bracket, we get;

$2\left(\frac{3}{4} x\right)+2(7)+(-3)\left(\frac{1}{2} x\right)+(-3)(-5)$

Hence, the expression $2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)$ is equivalent to $2\left(\frac{3}{4} x\right)+2(7)+(-3)\left(\frac{1}{2} x\right)+(-3)(-5)$

Thus, Option E is the correct answer.

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