Find the 11th term of this sequence -2, 10, -50

Mathematics

1 answer:

mrs_skeptik [129]4 years ago

8 0

Answer:

See below

Step-by-step explanation:

Formula

The general term we will use is

t = a r^(n - 1)

Givens

a = - 2

r = - 5

n = 11

t_n = a*r^(n - 1)

t_n = -2*(-5)^(11 - 1)

t_11 = -2*(-5)^10

t_11 = -2 (9,765,625)

t_11 = 19,531,250

Note

Before accepting this answer, you should check to see that the formula works with a value you know the answer to.

t_3 = -2(-5)^(3-1)

t_3 = -2(-5)^2

t_3 = -2 * 25

t_3 = - 50

Which is what your series shows.

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Find the indefinite integral by using the substitution x = 4 sec(θ). (Use C for the constant of integration.) x2 − 16 x d

guapka [62]

Answer:

$\frac{x^2-16}{2} + 16ln\frac{4}{x} +16C$

Step-by-step explanation:

Given the indefinite integral $\int\limits{\frac{x^2-16}{x} } \, dx$ , using the substitute

x = 4 sec(θ)...1

The integral can be calculated as thus;

First let us diffrentiate the substitute function with respect to θ

dx/dθ = 4secθtanθ

dx = 4secθtanθdθ... 2

Substituting equation 1 and 2 into the integral function we will have;

$\int\limits{\frac{(4sec \theta)^2-16}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\int\limits{\frac{16sec^2 \theta-16}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\int\limits{\frac{(16(sec^2 \theta-1)}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\\from \ trig \ identity,\ sec^2 \theta - 1 = tan^\theta\\\\\int\limits{\frac{16 tan^2 \theta}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\\\int\limits 16 tan^3 \theta d \theta\\\\$