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3 years ago

Given the sequence 38, 32, 26, 20, 14, ..., find the explicit formula.

Mathematics

2 answers:

-BARSIC- [3]3 years ago

8 0

Answer:

The explicit formula for the sequence is

Step-by-step explanation:

The above sequence is an arithmetic sequence

For an nth term in an arithmetic sequence

A(n) = a + ( n - 1)d

where a is the first term

n is the number of terms

d is the common difference

From the question

a = 38

d = 32 - 38 = - 6 or 20 - 26 = - 6 or

14 - 20 = - 6

So the formula for the sequence is

A(n) = 38 + ( n - 1)-6

= 38 - 6n + 6

We have the final answer as

Hope this helps you

bogdanovich [222]3 years ago

4 0

Answer:

$\huge\boxed{a_n=-6n+44}$

Step-by-step explanation:

This is an arithmetic sequence:

32 - 38 = -6

26 - 32 = -6

20 - 26 = -6

14 - 20 = -6

The common difference d = -6.

The explicit formula of an arithmetic formula:

$a_n=a_1+(n-1)(d)$

Substitute:

$a_1=38;\ d=-6$

$a_n=38+(n-1)(-6)$ <em>use the distributive property</em>

$a_n=38+(n)(-6)+(-1)(-6)\\\\a_n=38-6n+6\\\\a_n=-6n+(38+6)\\\\a_n=-6n+44$

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Juliette [100K]

Answer:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

Step-by-step explanation:

For this case we know that the currently mean is 2.8 and is given by:

$\bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8$

Where $w_i$ represent the number of credits and $X_i$ the grade for each subject. From this case we can find the following sum:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

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