Exercise 32 Page 712

We want to find the geometric means in the given sequence. To do so, we first have to find the value of the common ratio r. 6, ?,?, 162 From the sequence, we know the value of the first term and that there are 4 total terms. We can substitute 6 for a_1 and 4 for n in the general formula for the n^(th) term of a geometric sequence. a_n=a_1r^(n-1) ⇒ a_4= 6r^(4-1) We also know that the value of the fourth term is 162. We can substitute this in the above formula and solve for r.

a_4=6r^(4-1)

Substitute

a_4= 162

162=6r^(4-1)

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Solve for r

SubTerm

Subtract term

162=6r^3

DivEqn

.LHS /6.=.RHS /6.

27=r^3

RadicalEqn

sqrt(LHS)=sqrt(RHS)

sqrt(27)=r

CalcRoot

Calculate root

3=r

RearrangeEqn

Rearrange equation

r=3

We found that r here is 3. We will use this to calculate the geometric means we are looking for.