HELP ME OUT BRO PLEASE I HONESTLY DONT KNOW WHAT THIS QUESTION MEANS, I JUST DONT KNOW WHAT TO DO

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago

Find the center and radius of this circle x^2+y^2-6x-10y+30=0

Vsevolod [243]

Answer:

The circle's centre is at the position (3, 5), and it has a radius of 2

Step-by-step explanation:

First let's put it in a useful format by completing the squares:

x² + y² - 6x - 10y + 30 = 0

x² - 6x + y² - 10y = -30

x² - 6x + 9 + y² - 10y + 25 = -30 + 9 + 25

(x - 3)² + (y - 5)² = 4

This tells us that the centre position is (3, 5) and the radius is √4, or 2

5 0

3 years ago

Helppp!!!! please!!!

Ulleksa [173]

Answer:

$\boxed{Area = 32.5 mm^2}$

Step-by-step explanation:

Area of Triangle = $\frac{1}{2} (Base)(Height)$

Where base = 13 mm, Height = 5 mm

Area = 1/2 (13)(5)

=> Area = 1/2 (65)

=> Area = 32.5 mm^2

5 0

4 years ago

Read 2 more answers

Find the sum of the first four terms of the geometric sequence shown below. 4, 4/ 3, 4/9, ...

user100 [1]

It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is

$S_4=4+\dfrac43+\dfrac49+\dfrac4{27}$

It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.

Multiply the sum by 1/3 and you have

$\dfrac13S_4=\dfrac43+\dfrac49+\dfrac4{27}+\dfrac4{81}$

Now subtracting this from $S_4$ gives

$S_4-\dfrac13S_4=4-\dfrac4{81}$

That is, all the matching terms will cancel. Now solving for $S_4$ , you
have

$\dfrac23S_4=4\left(1-\dfrac1{81}\right)$
$S_4=6\left(1-\dfrac1{81}\right)$
$S_4=\dfrac{480}{81}=\dfrac{160}{27}$

3 0

4 years ago

Picture of the questions are below.

Zigmanuir [339]

Step-by-step explanation:

Look at the picture below

6 0

3 years ago

Read 2 more answers