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I'm extremely bad at doing these

I'm extremely bad at doing these-example-1
User Tim Stone
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2 Answers

10 votes

Answer:

Term 2 is 21, and term 3 is 147.

Explanation:

The geometric sequence is 3, 3r, 3r^2, 1029.


r = \sqrt[3]{ (1029)/(3) }


r = \sqrt[3]{343} = 7

So we have 3, 21, 147, 1029.

User Paulius Liekis
by
8.7k points
7 votes

The first few terms of a geometric sequence with first term
a and common ratio
r look like


a, ar , ar^2, ar^3, \ldots

and so on. Notice that the
n-th term (where
n is a natural number) is
ar^(n-1).

For this particular sequence, the first term is


a=3

and the fourth term is


ar^(4-1) = ar^3 = 1029

Substitute
a=3 into the second equation and solve for
r.


3r^3 = 1029 \implies r^3 = 343 \implies r = \sqrt[3]{343} = \sqrt[3]{7^3} = 7

Then the two terms between the 1st and 4th - i.e. the 2nd and 3rd terms - are


ar = 3*7 = \boxed{21}

and


ar^2 = 3*7^2 = \boxed{147}

User Alassane Ndiaye
by
8.6k points

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