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Vitek1552 [10]
3 years ago
13

Gouilleul Sequence.

Mathematics
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

Step-by-step explanation:

No idea what a gouilleul Sequence is, but hopefully it will help to show how I would find the answer.

A geometric  sequence means a term is multiplied by something over and over So say we start with 100 and the common ratio is .5 then the sequence would go 100,50,25,12.5,6.25 and so on.  Another way of writing it is:

100*.5^0 = 100

100*5^1 = 50

100*.5^2 = 100*.25 = 25

and so on.  

ANyway, here we have the first and fourth term.  so that's 3.5*x^0 and 3.5*x^3.  Keep in mind the exponent will be n-1 when the term you want to find is the nth term.  Anyway, we know 3.5*x^3 = 28, so we just use algebra to find x, which is the common ratio.

3.5*x^3 = 28

x^3 = 8

x = 2

So the common ratio is 2, or in other words it keeps doubling.  You could try it real quick, keep doubling the starting term 3.5

3.5, 7, 14, 28 and 28 is the fourth term.  

So we know the common ratio.  you could just add the four terms listed then for part b, but there is an equation, for in case they ask you to find the sum of say the first 100 terms.  that wouldn't be fun.  Anyway, the formula is a((1-r^n)/(1-r)) where a is the starting term, r is the common ratio and n is the number term.  In your case that's 3.5((1-2^4)/(1-2)) = 52.5.  You can actually check since it's only 4 terms.  3.5 + 7 + 14 + 28 = 52.5.

If you want to know how that formula was gotten let em know and I can explain or give a link to a video or something to show.

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18.95

Step-by-step explanation:

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Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on
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<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest (I_1) earned in first account for one year,

\text {simple interest}=\frac{\text {pnr}}{100}

Where  

p = amount invested in first account

n = number of years  

r = rate of interest

hence, by using above equation we get (I_1) as,  

I_{1}=\frac{P \times 1 \times 3}{100} ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest (I_2) earned in second account,

I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

I = I_1+ I_2 ---- eqn 4

By substituting eqn 1 , 2, 3 in eqn 4

\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}

\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400  

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600  

Hence, Mathew invested $600 and $2400 in each account.

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Answer:

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Step-by-step explanation:

<u><em>Explanation:-</em></u>

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