Find the value of the 60th term of the following sequence: 33, 28, 23, 18

Mathematics

1 answer:

Otrada [13]3 years ago

4 0

Step-by-step explanation:

The first term of the AP i.e a = 33

and common difference i.e d = 28-33 = -5

So, 60th term = a +(n-1)d = a+ 59d

= 33 + 59( -5)

= 33- 295

= - 262

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I'm studying a new bacteria that doubles in numbers every 25 minutes. If i start with 50 bacteria, how long until i have 5 milli

dimaraw [331]

Answer:

415.63 minutes

Step-by-step explanation:

Growth can be represented by the equation $A=A_0e^{rt}$ . We can find the rate at which it grows by using t=25 minutes and $\frac{A_{0}}{A} =2$ or double the amount at that time. The first step we always take is to divide $A_0$ by A.

$[tex]\frac{A_{0}}{A}=e^{rt}\\2=e^{r(25)}$

$2=e^{(25)r}$

To solve for r, we will take the natural log of both sides and use log rules to isolate r.

$ln 2=ln e^{(25)r}\\ln 2=25r (ln e)\\\frac{ln2}{25} =r$

We know $lne=1$ so we were able to cancel it out and divide both sides by 25.

We solve with a calculator $\frac{ln2}{25} =r\\0.0277=r$

We change 0.0277 into a percent by multiplying by 100 to get 2.77% as the rate.

The equation is $A=A_0e^{0.0277t}$ .

We repeat the step above substituting A=5,000,000, $A_0$ =50, and r=0.02777. Then solve for t.

$5000000=50e^{0.0277t}\\\frac{5000000}{50} =e^{0.0277t}\\100000=e^{0.0277t}\\ln100000=lne^{0.0277t}\\ln100000=0.02777t(lne)\\\frac{ln100000}{0.0277} =t$