Help please

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago

Please help me ASAP. Find the area of the figure

ValentinkaMS [17]

Answer:

55.8

Step-by-step explanation:

all ya gotta do is multiply the numbers together

8 0

3 years ago

Based on these results, if 250 children are asked about their favorite activity, how many children will prefer organize sports?

ser-zykov [4K]

<u>Answer:</u>

60 out of 250 children will prefer organized sports.

<u>Step-by-step explanation:</u>

We are given the results of a random survey of children between the ages of 13 and 18 about their favorite activity.

Based on these results, we are to find the number of children who will prefer organized sports if 250 children were asked about their favorite activity.

Children who will prefer organized sports = $\frac{37}{155} \times 250$ = 60

5 0

3 years ago

Triangle HAM is translated using the rule (x,y)→(x+6, y) to create triangle H'A'M'.

Andrej [43]

The statement would best describe the line segments drawn would be they are parallel and congruent. The correct answer between all the choices given is the second choice or letter B. I am hoping that this answer has satisfied your query about and it will be able to help you.

4 0

3 years ago

Read 2 more answers

What is the missing number in the boxes

snow_tiger [21]

(311&1/4-186&3/4)/(5-3)=62&1/4
so the difference between each adjacent number is 62&1/4
when the top number is 4, the bottom number is 186&3/4+62&1/4=249 (a)
when the top number is 4, the bottom number is 311&1/4+62&1/4=373&1/2 (b)
C=10

5 0

3 years ago