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chubhunter [2.5K]
2 years ago
11

What is a formula for the nth term of a given sequence

Mathematics
2 answers:
Sedbober [7]2 years ago
7 0

Answer:

a_{n} = -12-4(n-1)

Step-by-step explanation:

We have given a arithmetic sequence.

-12,-16,-20,...

We have to find formula for a given sequence.

The general formula for nth term of sequence is :

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

In given sequence,

a_{1} = -12

d is the common difference between consecutive terms.

d = -16-(-12)  = -16+12

d = -4

Putting given values in formula, we have

a_{n} = -12-4(n-1) which is the answer.

IceJOKER [234]2 years ago
5 0

Answer:

a_n=-12-4(n-1)

Step-by-step explanation:

The given sequence is

-12,-16,-20...

The first term of this sequence is a_1=-12.

The common difference is

d=-16--12

d=-16+12=-4

The nth term of this arithmetic sequence is;

a_n=a_1+d(n-1)

We substitute the values for the first term and the common difference to obtain;

a_n=-12-4(n-1)

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Ilia_Sergeevich [38]

Answer:

\overline{v}_{@\Delta t=0.01s}=-15.22ft/s, \overline{v}_{@\Delta t=0.005s}=-15.11ft/s, \overline{v}_{@\Delta t=0.002s}=-15.044ft/s, \overline{v}_{@\Delta t=0.001s}=-15.022ft/s

Step-by-step explanation:

Now, in order to solve this problem, we need to use the average velocity formula:

\overline{v}=\frac{y_{f}-y_{0}}{t_{f}-t_{0}}

From this point on, you have two possibilities, either you find each individual y_{f}, y_{0}, t_{f}, t_{0} and input them into the formula, or you find a formula you can use to directly input the change of times. I'll take the second approach.

We know that:

t_{f}-t_{0}=\Delta t

and we also know that:

t_{f}=t_{0}+\Delta t

in order to find the final position, we can substitute this final time into the function, so we get:

y_{f}=29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}

so we can rewrite our formula as:

\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-y_{0}}{\Delta t}

y_{0} will always be the same, so we can start by calculating that, we take the provided function ans evaluate it for t=1s, so we get:

y_{0}=29t-22t^{2}

y_{0}=29(1)-22(1)^{2}

y_{0}=7ft

we can substitute it into our average velocity equation:

\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-7}{\Delta t}

and we also know that the initil time will always be 1, so we can substitute it as well.

\overline{v}=\frac{29(1+\Delta t)-22(1+\Delta t)^{2}-7}{\Delta t}

so we can now simplify our formula by expanding the numerator:

\overline{v}=\frac{29+29\Delta t-22(1+2\Delta t+\Delta t^{2})-7}{\Delta t}

\overline{v}=\frac{29+29\Delta t-22-44\Delta t-22\Delta t^{2}-7}{\Delta t}

we can now simplify this to:

\overline{v}=\frac{-15\Delta t-22\Delta t^{2}}{\Delta t}

Now we can factor Δt to get:

\overline{v}=\frac{\Delta t(-15-22\Delta t)}{\Delta t}

and simplify

\overline{v}=-15-22\Delta t

Which is the equation that will represent the average speed of the ball. So now we can substitute each period into our equation so we get:

\overline{v}_{@\Delta t=0.01s}=-15-22(0.01)=-15.22ft/s

\overline{v}_{@\Delta t=0.005s}=-15-22(0.005)=-15.11ft/s

\overline{v}_{@\Delta t=0.002s}=-15-22(0.002)=-15.044ft/s

\overline{v}_{@\Delta t=0.001s}=-15-22(0.001)=-15.022ft/s

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Step-by-step explanation:

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