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Soloha48 [4]
3 years ago
6

Which of the following geometric series converges?

Mathematics
1 answer:
Mars2501 [29]3 years ago
8 0
All three series converge, so the answer is D.
The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.
Consider a geometric sequence with the first term a and common ratio |r| < 1. Then the n-th partial sum (the sum of the first n terms) of the sequence is

Multiply both sides by r :

Subtract the latter sum from the first, which eliminates all but the first and last terms:

Solve for :

Then as gets arbitrarily large, the term will converge to 0, leaving us with

So the given series converge to
(I) -243/(1 + 1/9) = -2187/10
(II) -1.1/(1 + 1/10) = -1
(III) 27/(1 + 1/3) = 18
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Answer: Option 'A' is correct.

Step-by-step explanation:

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7 0
3 years ago
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andrew-mc [135]

Answer:

Multiply vector c by the scalar -1/2.

Step-by-step explanation:

Look at vector c.

It has an x component of 4 and a y component of 4.

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c = 4i + 4j

Now look at vector d, and write it also as a sum of its x and y components.

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That worked. By multiplying vector c by the scalar -1/2, you end up with vector d.

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2 years ago
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