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Murrr4er [49]
3 years ago
14

Write the explicit formula for the geometric sequence. a1 = -5 a2 = 20 a3 = -80

Mathematics
1 answer:
Anni [7]3 years ago
6 0
a_1 = -5;\ a_2 = 20;\ a_3 = -80 \\\\a_2:a_1=r\ and\ a_3:a_2=r\\\\r=20:(-5)=-4\ check\ r=-80:20=-4\\\\a_n=a_1r^{n-1}\to a_n=-5\cdot(-4)^{n-1}=-5\cdot(-4)^{-1}\cdot(-4)^n\\\\=-5\cdot\left(-\dfrac{1}{4}\right)\cdot(-4)^n=\dfrac{5}{4}\cdot(-4)^n\\\\Answer:\boxed{a_n=-5\cdot(-4)^{n-1}\ other\ a_n=\frac{5}{4}\cdot(-4)^n}
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777dan777 [17]

Answer:

The first term is 13/3 and the common difference is d = 1/12

The formula is a(n) = 13/3 + (1/12)(n - 1)        

Step-by-step explanation:

The general equation for an arithmetic progression is:

a(n) = a(1) + d(n - 1), where d is the common difference/

Case 1:  n = 7:  29/6 = a(1) + d(7 - 1), or 29/6 = a(1) + d(6)

Case 2:  n = 15:  11/2 = a(1) + d(15 - 1) =  a(1) + d(14)

Then our system of linear equations is:

a(1) + 6d = 29/6

a(1) + 14d = 11/2

Let's solve this by elimination by addition and subtraction.  Subtract the first equation from the second.  We get:

                                                             

Substituting 1/12 for d in the first equation, we get:

a(1) + 14(1/12) = 11/2 or 66/12 (using the LCD 12)

Then a(1) = 66/12 - 14/12 = 52/12 = 13/3

The first term is 13/3 and the common difference is d = 1/12

The arithmetic sequence formula for this problem is thus:

a(n) = 13/3 + (1/12)(n - 1)                                                                          8

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3 years ago
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