(3/5)^-2

In an arithmetic sequence, the nth term an is given by the formula an=a1+(n−1)d, where a1 is the first term and d is the commo

Dmitry_Shevchenko [17]

Answer:

$a_{10} = \frac{10}{65536}$

Step-by-step explanation:

The first step to solving this problem is verifying if this sequence is an arithmetic sequence or a geometric sequence.

This sequence is arithmetic if:

$a_{3} - a_{2} = a_{2} - a_{1}$

We have that:

$a_{3} = 40, a_{2} = 10, a_{3} = \frac{5}{2}$

$a_{3} - a_{2} = a_{2} - a_{1}$

$\frac{5}{2} - 10 = 10 - 40$

$\frac{-15}{2} \neq -30$

This is not an arithmetic sequence.

This sequence is geometric if:

$\frac{a_{3}}{a_{2}} = \frac{a_{2}}{a_{1}}$

$\frac{\frac{5}[2}}{10} = \frac{10}{40}$

$\frac{5}{20} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$

This is a geometric sequence, in which:

The first term is 40, so $a_{1} = 40$

The common ratio is $\frac{1}{4}$ , so $r = \frac{1}{4}$ .

We have that:

$a_{n} = a_{1}*r^{n-1}$

The 10th term is $a_{10}$ . So:

$a_{10} = a_{1}*r^{9}$

$a_{10} = 40*(\frac{1}{4})^{9}$

$a_{10} = \frac{40}{262144}$

Simplifying by 4, we have:

$a_{10} = \frac{10}{65536}$

3 0

3 years ago

How do you solve the system 3x-y z=5, x 3y 3z=-6, and x 4y-2x=12?

dlinn [17]

3x - y + z = 5 . . . (1)
x + 3y + 3z = -6 . . . (2)
x + 4y - 2z = 12 . . . (3)

From (2), x = -6 - 3y - 3z . . . (4)
Substituting for x in (1) and (3) gives
3(-6 - 3y - 3z) - y + z = 5 => -18 - 9y - 9z - y + z = 5 => -10y - 8z = 23 . . (5)
-6 - 3y - 3z + 4y - 2z = 12 => y - 5z = 18 . . . (6)

(6) x 10 => 10y - 50z = 180 . . . (7)
(5) + (7) => -58z = 203
z = 203/-58 = -3.5

From (6), y - 5(-3.5) = 18 => y = 18 - 17.5 = 0.5
From (4), x = -6 - 3(0.5) - 3(-3.5) = -6 - 1.5 + 10.5 = 3

x = 3, y = 0.5, z = -3.5

8 0

4 years ago

DVD Video Rentals (Refer to Example 3.) The func-

Neporo4naja [7]

Answer:

$(a)\ V = \{(N,30),(R,37),(S,17)\}$