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2 years ago

In an arithmetic sequence, = 5 and = 11.

Mathematics

1 answer:

Free_Kalibri [48]2 years ago

5 0

Answer:

1134.62 (THE SUM OF 20 TERMS)

Step-by-step explanation:

First we write out our given information:

a8=2a2

a11=18

an is an arithmetic sequence.

Where here an means the nth term of our sequence.

What does an arithmetic sequence mean? It means to get to the next term in your sequence you add a constant (c) each time:

an+1=an+c

Equivalently:

an+1−an(n+1)−n=c

So an is of slope c (c2 is another constant):

an=cn+c2

Where here c2=a0 (Substitute in n=0 and see why that has to be the case if we let a0 exist)

Now we use the other given information to try to come up with a solution.

Let n=2:

a2=2c+c2

Let n=8, using the above equation we have:

a8=8c+c2=2a2=4c+2c2

Let n=11

a11=18

a11=11c+c2

But a11−a8=(11c+c2)−(8c+c2)=3c

Hence, a11=3c+a8

a11=3c+4c+2c2=18

a11=3c+8c+c2=18

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Given z1 and z2 below. Calculate z1/z2 and write your answer in rcistheta form.

andre [41]

Answer:

(2/9)cis(30°)

Step-by-step explanation:

We can express the two numbers in magnitude∠angle form, then find their ratio.

|z1| = 3√((-1)² +(√3)²) = 3√4 = 6

∠z1 = arctan((3√3)/(-3)) = -arctan(√3) = 120°

z2 = 27∠90°

So, the ratio is ...

z1/z2 = (6∠120°)/(27∠90°) = (6/27)∠(120°-90°)

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2 years ago

Match each series with the equivalent series written in sigma notation

PIT_PIT [208]

Answer:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

$3 + 12 + 48 + 192 + 768$

The expression can be written as:

$3 \to 3*4^{0$ --- 0

$12 \to 3 * 4^{1$ ---- 1

$48 \to 3 * 4^{2$ --- 2

$192 \to 3 * 4^{3$ ---- 3

$768 \to 3 * 4^{4$ ---- 4

For the nth term, the expression is:

$Term = 3 * 4^{n$ ---- n

So, the summation is:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

Solving (b):

$4 + 32 + 256 + 2048 + 16384$

The expression can be written as:

$4 \to 4 * 8^0$ --- 0

$32 \to 4 * 8^1$ ---- 1

$256 \to 4 * 8^2$ --- 2

$2048 \to 4 * 8^3$ ---- 3

$16384 \to 4 * 8^4$ ---- 4

For the nth term, the expression is:

$Term \to 4 * 8^n$ ---- n

So, the summation is:

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

Solving (c):

$2 + 6 + 18 + 54 + 162$

The expression can be written as:

$2 \to 2 * 3^0$ --- 0

$6 \to 2 * 3^1$ ---- 1

$18 \to 2 * 3^2$ --- 2

$54 \to 2 * 3^3$ ---- 3

$162 \to 2 * 3^4$ ---- 4

For the nth term, the expression is:

$Term \to 2 * 3^n$ ---- n

So, the summation is:

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

Solving (d):

$3 + 15 + 75 + 375 + 1875$

The expression can be written as:

$3 \to 3 * 5^0$ --- 0

$15 \to 3 * 5^1$ ---- 1

$75 \to 3 * 5^2$ --- 2

$375 \to 3 * 5^3$ ---- 3

$1875 \to 3 * 5^4$ ---- 4

For the nth term, the expression is:

$Term \to 3 * 5^n$ ---- n

So, the summation is:

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

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Answer:

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Step-by-step explanation:

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