Help with 2 Algebra 2 questions (I always choose best answers personally)
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Help with 2 Algebra 2 questions (I always choose best answers personally)

[From: ] [author: ] [Date: 12-03-25] [Hit: ]
(a+2b)/3,2. Since any geometric sequence is formed by multiplying one term by a constant called the common ratio, r,-2, -2r,......
I know it seems like i'm asking a load of questions but I really need help.

1) Find the missing two term in the arithmetic sequence: a,_,_,b

2) Insert two terms between -2 and 128, if this is a geometric sequence

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1) Since any arithmetic sequence is formed by adding a constant called the common difference, d, to one term to obtain the next term in the sequence, the sequence is:

a, a+d, a+2d, a+3d=b

and the common difference is:
d = (b-a)/3

Starting with the first term a, and adding the common difference, (b-a)/3 to each term to get the next term in the sequence:

a, a+(b-a)/3, a+2(b-a)/3, b

or, simplifying, the sequence becomes:

a, (2a+b)/3, (a+2b)/3, b

2. Since any geometric sequence is formed by multiplying one term by a constant called the common ratio, r, the sequence is:

-2, -2r, -2r², -2r³ = 128

Ignoring the complex roots of the cubic, the common ratio is:
r³ = -64
r = -4

Starting with the first term, -2, and multiplying each term by the common ratio, -4, to get the next term, the sequence becomes:

-2, -2(-4), -2(-4)², 128
or
-2, 8, -32, 128

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1. a,b,a,b
2. 30.5,63
1
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