Hello? Help me Please?

Question

determine the common ratio, the eight term, and a formua for the nth term of each geometric sequence.
a. 1,-2,-4,-8,16.....

1. the fourth and seventh terms of an arithmetic progression are -8,and -4 respectively. determine the first term and a formula for the nth term.

Answer 1

I think -4 should be +4 in your first problem. It's an alternating series with common ratio -2 (answer).

You show 5 terms, so the eighth is easy:

1, -2, 4, -8, 16, -32, 64, -128 (8th term, answer)

nth term g(n) = (-1)^(n-1) 2^(n-1) = (-2)^(n-1) (answer)


Next problem

General term is a(n) = a(1) + (n-1)d

where d is common difference. You have

a(4) = a(1) + 3d = -8 (Eq 1)
a(7) = a(1) + 6d = -4 (Eq 2)

6d - 3d = 3d = -4 - (-8) = 4 ==> d = 4/3

so from Eq 1, a(1) + 3(4/3) = -8
a(1) = -8 - 4 = -12 (answer)

The nth term is

a(n) = -12 + (n - 1)(4/3)
a(n) = -12 - 4/3 + 4n/3
a(n) = (4n - 40)/3 = 4/3 (n - 10) (answer)

Check: a(1) = 4/3 (-9) = -12 (OK)
a(4) = 4/3 (-6) = -8 (OK)
a(7) = 4/3 (-3) = -4 (OK)

These answers are right.

Answer 2

17

Answer 3

are you sure you don't mean

1,-2,4,-8,16 if so then

(-2)^(n - 1)

8th Term = (-2)^(8 - 1) = (-2)^7 = -128

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a, b, c, -8, d, e, -4

an = a1 + (n - 1)d

a(4) = a1 + (4 - 1)d
-8 = a1 + 3d
a1 = -3d - 8

a(7) = a1 + (7 - 1)d
-4 = a1 + 6d
a1 = -6d - 4

-3d - 8 = -6d - 4
3d = 4
d = (4/3)

a1 = -6(4/3) - 4
a1 = (-24/3) - 4
a1 = -8 - 4
a1 = -8 + (-4)
a1 = -12

First Term : -12
Formula an = (4/3)(n - 1) - 12

1st = -12
2nd = -10 2/3
3rd = -9 1/3
4th = -8
5th = -6 2/3
6th = -5 1/3
7th = -4