All categories

4 years ago

Which are the first five terms of a geometric sequence in which the second term is 6 and the fourth term is 54?

1 answer:

7 0

now, let's recall that a geometric sequence is one that uses some "r" common ratio to get the next term, by simply multiplying the current term by it.

$\bf \begin{array}{|cl|ll} \cline{1-2} term&value\\ \cline{1-2} a_1&\underline{\qquad }\\&\\ a_2&6\\&\\ a_3&\underline{6(r)}\\&\\ a_4&6(r)(r)\\&\\ &54\\ \cline{1-2} \end{array}\qquad \implies \begin{array}{llll} 54=6r^2\implies \cfrac{54}{6}=r^2\implies 9=r^2\\\\ \sqrt{9}=r\implies 3=r \end{array} \\\\[-0.35em] ~\dotfill\\\\ a_1=6\div 3\implies a_1=2~\hfill a_3=6(3)\implies a_3=18$

and of course, the next term or a₅ = 54(3) --> a₅ = 162.

You might be interested in

Can someone help me with algebra 2 with an explanation?

sammy [17]

Answer: This took me a long time to type, hope you read it and find it helpful and easy to understand...

1. $x=1$

2. $x=\frac{3}{2}$

3. $x=1$

4. $x=\frac{6}{5}$

Step-by-step explanation:

$\frac{1}{x}-\frac{x-2}{3x}=\frac{4}{3x}$

To make it easier to add / subtract fractions, we are looking for common denominators. I can see that if I move the fraction on the left with a denominator 3x to the right, I'll be able to add two equations with like denominator.

Let's add $\frac{x-2}{3x}$

$\frac{1}{x}=\frac{4}{3x}+\frac{x-2}{3x}$

Add the numerators and keep the same denominator.

$\frac{1}{x}=\frac{4+x-2}{3x}$

$\frac{1}{x}=\frac{2+x}{3x}$

To get rid of the denominator x on the left side, we can multiply by x.

$(x)\frac{1}{x}=\frac{2+x}{3x}(x)$

$1=\frac{2+x}{3}$

Now multiply by 3.

$(3)1=\frac{2+x}{3}(3)$

$3=2+x$

Subtract 2.

$3-2=x\\1=x$

The value of x is 1.

Proof.

$\frac{1}{x}-\frac{x-2}{3x}=\frac{4}{3x}$

$\frac{1}{1}-\frac{1-2}{3(1)}=\frac{4}{3(1)}$

$1-\frac{-1}{3}=\frac{4}{3}$

$1+\frac{1}{3}=\frac{4}{3}$

$\frac{1*3+1}{3} =\frac{4}{3}\\\frac{4}{3}=\frac{4}{3}$

I can't show the proof to the following problems because I have a 5000 character limit.

-----------------------------------------------------------------------------------------

$\frac{5x-5}{x^2-4x} -\frac{5}{x^2-4x}=\frac{1}{x}$

Again, we have like denominators, therefore, we can simply subtract the numerators and keep the same denominator.

$\frac{5x-5-5}{x^2-4x}=\frac{1}{x}$

$\frac{5x-10}{x^2-4x}=\frac{1}{x}$

Again, we can multiply by x to get rid of the x in the denominator.

$(x)\frac{5x-10}{x^2-4x}=\frac{1}{x}(x)$

Do not distribute the x in the numerator yet because we're gonna eliminate it later.

$\frac{(x)(5x-10)}{x^2-4x}=1$

Factor the denominator.

$\frac{(x)(5x-10)}{(x)(x-4)}=1$

Simplify x's.

$\frac{5x-10}{x-4} =1$

Multiply by x-4 to get rid of the denominator.

$(x-4)\frac{5x-10}{x-4} =1(x-4)$

$5x-10=x-4$

Add 10

$5x=x-4+10$

Subtract x

$5x-x=-4+10$

Combine like terms;

$4x=6$

Divide by 4.

$\frac{4x}{4}=\frac{6}{4}$

$x=\frac{6}{4}$

Simplify by 2.

6/2 = 3

4/2 = 2

$x=\frac{3}{2}$

-----------------------------------------------------------------------

$\frac{x^2-7x+10}{x}+\frac{1}{x}=x+4$

Same denominator, add numerators.

$\frac{x^2-7x+10+1}{x}=x+4$

$\frac{x^2-7x+11}{x}=x+4$

Multiply by x to get rid of the x.

$(x)\frac{x^2-7x+11}{x}=(x+4)(x)$

$x^2-7x+11=x^2+4x$

Subtract $-x^2-4x$

$x^2-7x+11-x^2-4x=0$

Combine like terms;

$-11x+11=0$

Subtract 11.

$-11x=-11$

Divide by -11

$x=\frac{-11}{-11} \\x=1$

-------------------------------------------------------------------------

$\frac{x^2+7x+10}{5x-30}+\frac{x}{x-6}=\frac{x^2-13x+40}{5x-30}$

It's easier if you move the operation with denominator x+6 to the right side with negative sign and bring the operation on the right side to the left side with negative sign as well.

$\frac{x^2+7x+10}{5x-30}-\frac{x^2-13x+40}{5x-30}=-\frac{x}{x-6}$