Find the 53rd term of the arithmetic sequence 27,11, -5

Mathematics

1 answer:

vivado [14]2 years ago

3 0

Answer:

The 53rd term of this arithmetic sequence is -805.

Step-by-step explanation:

The general rule of an arithmetic sequence is the following:

$a_{n+1} = a_{n} + d$

In which d is the common diference between each term, that is, $d = a_{3} - a_{2} = a_{2} - a_{1}$ .

To find the nth term of the sequence, this equation can be written as:

$a_{n} = a_{1} + (n-1)d$

27,11, -5

So $a_{1} = 27, a_{2} - a_{1} = 11 - 27 = -16[/tex[tex]a_{n} = a_{1} + (n-1)d$

$a_{53} = a_{1} + (52)d = 27 + 52*(-16) = -805$

The 53rd term of this arithmetic sequence is -805.

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gayaneshka [121]

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2 years ago

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$