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2 years ago

Find the 8th term of the geometric sequence 7,−21,63,

Mathematics

2 answers:

finlep [7]2 years ago

5 0

Answer:

8th term is -15309

Step-by-step explanation:

${ \boxed{ \bf{u_{n} = a( {r}^{n - 1} ) }}} \\ { \tt{u_{8} = 7( {( - 3)}^{8 - 1}) }} \\ { \tt{u_{8} = 7( - 2187)}} \\ { \tt{u _{8} = - 15309}}$

r is the common difference, r = -21/7 = -3

noname [10]2 years ago

5 0

Answer:

a₈ = - 15309

Step-by-step explanation:

The nth term of a geometric sequence is

$a_{n}$ = a₁ $(r)^{n-1}$

where a₁ is the first term and r the common ratio

Here a₁ = 7 and r = $\frac{a_{2} }{a_{1} }$ = $\frac{-21}{7}$ = - 3 , then

a₈ = 7 × $(-3)^{7}$ = 7 × - 2187 = - 15309

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Answer:

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Step-by-step explanation:

For this case we have the following equation given:

$y= x^2 -6x +14$

We can complete the square like this:

$y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2$

And solving we have:

$y= x^2 -6x +9 + 14 -9$

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And we can write the expression like this:

$y-5 = (x-3)^2$

The vertex for this case would be:

$V= (3,5)$

And the minimum for the function would be 3 and there is no maximum value for the function

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