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
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
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
2. 30.5,63