All categories

2 years ago

What is the fourth term of the sequence? a1 = m an = 2an-1

Mathematics

2 answers:

mrs_skeptik [129]2 years ago

5 0

Answer:

Fourth term of the sequence is 8m

Step-by-step explanation:

Given: Sequence $a_1=m \ and\ a_n=2a_{n-1}$

We have to find the fourth term of the given sequence.

Consider the given sequence $a_n=2a_{n-1}$

Forth term will be $a_4=2a_3$ ....(1)

We find third term first

$a_3=2a_2$ ...(2)

Also, given $a_1=m$

and $a_2=2a_1=2m$ .....(3)

Put (3) in (2), we get,

$a_3=2(2m)=4m$ ...(4)

Put (4) in (1) , we get,

$a_4=2a_3=2(4m)=8m$

Thus, Fourth term of the sequence is 8m

natita [175]2 years ago

4 0

A2=2a1=2m; a3=2a2=4m; a4=2a3=8m.

You might be interested in

A room with a length of 270 cm and width 150 cm is to be covered with square tiles .What is the largest size of tiles to be used

Damm [24]

Find the greatest common factor of both numbers:

Factors for 150:

150: 1, 2, 5 , 6, 10, 15, 25, 30, 50, 75, 150

Factors for 270:

1, 2, 3, 5, 6 , 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270

The largest number they both have in common is 30, so the largest tile could be 30 cm^2.

5 0

2 years ago

A gallon of paint covers 400 square feet. How many square feet will 2 3/8 gallons of paint cover?

Black_prince [1.1K]

1 gal = 400 ft²
2 3/8 gal = ? ft²

to solve, use similar ratios with units gal/ft² and compare

(1)/(400) = (2 3/8)/(x)

*multiply both sides by (x) and divide both sides by (1)/(400), you get:

x = (400)*(2 3/8)/(1) = 400*(2 3/8) = 400*2 + 400*3/8 = 800 + 150 = 950 ft²

answer is 950 ft²

8 0

2 years ago

Brenda earns $1,700 per month after taxes. She is working on her budget and has the first three categories finished.

REY [17]

Answer:

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Another path in the community is 2.1 miles long. It has benches at 1/3 And 2/3 of the distance from beginning to end of the path

emmasim [6.3K]

Answer:

bench at 1/3 is at 0.7 miles

bench at 2/3 is at 1.4 miles

Step-by-step explanation:

Given

Path length, x= 2.1 miles

It has benches at 1/3 And 2/3 of the distance from beginning to end of the path

1/3 of the distance from beginning to end of the path= 1/3x

Putting value of x in above we get:

=1/3(2.1)

=2.1/3

=0.7

The bench at 1/3 of the distance is at 0.7 miles from the beginning to end of path

2/3 of the distance from beginning to end of the path= 2/3x

Putting value of x in above we get:

=2(2.1)/3

=4.2/3

=1.4

The bench at 2/3 of the distance is at 1.4 miles from the beginning to end of path!

5 0

2 years ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago