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valentina_108 [34]
3 years ago
8

Find a formula for the general term an of the sequence, assuming that the pattern of the first few terms continues. (Assume that

n begins with 1.)
{1, 7, 13, 19, 25, . . . }
Mathematics
1 answer:
s2008m [1.1K]3 years ago
4 0

Answer:

the formula is a_n = 6n - 5

Step-by-step explanation:

The formula is

Given that

{1, 7, 13, 19, 25, . . . }

The difference between the numbers is

7 -  1 = 6

13 -7 = 6

19 - 13 = 6

a = 1

d = 6

Now the formula is

a_n = a + (n - 1)d

= 1 + (n - 1)6

= 1 + 6n - 6

= 6n - 5

So, the formula is a_n = 6n - 5

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At the beginning of each of her four years in college, Miranda took out a new Stafford loan. Each loan had a principal of $5,500
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Answer:

D. $31,337.27

Step-by-step explanation:

We have that the initial amount of the loan is $5500.

Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.

The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.

Also, it is given that the loan was compounded annually.

We have the formula as,

P=\frac{\frac{r}{n}\times PV}{1-(1+\frac{r}{n})^{-t\times n}}

i.e. PV=\frac{P\times [1-(1+\frac{r}{n})^{-t\times n}]}{\frac{r}{n}}

Substituting the values, we get,

i.e. PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}

i.e. 22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}

i.e. 22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}

i.e. 22000=\frac{P\times [1-0.4735]}{0.00625}

i.e. 22000=\frac{P\times 0.5265}{0.00625}

i.e. P=\frac{22000\times 0.00625}{0.5265}

i.e. P=\frac{137.5}{0.5265}

i.e. P=261.16

Thus, the total lifetime cost to pay of the loans compounded annually  = 261.16 × 120 = $31,339.2

Hence, the total cost close to the answer is $31,337.27

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Step-by-step explanation:

Formula for standard error :

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As per given , we have

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