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3 years ago

Which term of the Ap: 3,8,13,18.......is 78?

Mathematics

1 answer:

Alex Ar [27]3 years ago

4 0

Answer:

$\bold{a_{16}=78}$

Step-by-step explanation:

$a_1=3,\ a_2=3+5=8,\ a_3=8+5=13,\ a_4=13+5=18\ \implies\ d=5\\\\a_n=a_1+(n-1)d\\\\a_n=3+(n-1)(5)\\\\a_n=3+5n-5\\\\a_n=5n-2\\\\ a_n=78\quad\implies\quad n=\,??\\\\78=5n-2\\\\78+2=5n\\\\5n=80\\\\n=16$

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The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola that describes the path of her jump is $\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

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$\mathbf{Length = 28}$

<em>Assume she starts from the origin (0,0)</em>

The midpoint would be:

$\mathbf{Mid = \frac 12 \times Length}$

$\mathbf{Mid = \frac 12 \times 28}$

$\mathbf{Mid = 14}$

So, the vertex of the parabola is:

$\mathbf{Vertex = (Mid,Height)}$

Express properly as:

$\mathbf{(h,k) = (14,20)}$

A point on the graph would be:

$\mathbf{(x,y) = (28,0)}$

The equation of a parabola is calculated using:

$\mathbf{y = a(x - h)^2 + k}$

Substitute $\mathbf{(h,k) = (14,20)}$ in $\mathbf{y = a(x - h)^2 + k}$

$\mathbf{y = a(x - 14)^2 + 20}$

Substitute $\mathbf{(x,y) = (28,0)}$ in $\mathbf{y = a(x - 14)^2 + 20}$

$\mathbf{0 = a(28 - 14)^2 + 20}$

$\mathbf{0 = a(14)^2 + 20}$

Collect like terms

$\mathbf{a(14)^2 =- 20}$

Solve for a

$\mathbf{a =- \frac{20}{14^2}}$

$\mathbf{a =- \frac{20}{196}}$

Simplify

$\mathbf{a =- \frac{5}{49}}$

Substitute $\mathbf{a =- \frac{5}{49}}$ in $\mathbf{y = a(x - 14)^2 + 20}$

$\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

Hence, the equation of the parabola that describes the path of her jump is $\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

See attachment for the graph

Read more about equations of parabola at:

brainly.com/question/4074088

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2 years ago

Which term of the Ap: 3,8,13,18.......is 78?​

Which term of the Ap: 3,8,13,18.......is 78?