Answer :

We are given the first term as
[tex]$$
a = 3125,
$$[/tex]
and the fourth term as
[tex]$$
b = 32.
$$[/tex]
We wish to insert two geometric means (GM[tex]\(_1\)[/tex] and GM[tex]\(_2\)[/tex]) between them so that the four-term sequence
[tex]$$
3125,\quad GM_1,\quad GM_2,\quad 32
$$[/tex]
forms a geometric progression. In a geometric progression the terms satisfy
[tex]$$
a,\quad ar,\quad ar^2,\quad ar^3 = b,
$$[/tex]
where [tex]\( r \)[/tex] is the common ratio.

### Step 1. Find the Common Ratio

Since the fourth term is given by
[tex]$$
ar^3 = 32,
$$[/tex]
we substitute [tex]\( a = 3125 \)[/tex] to obtain
[tex]$$
3125 \cdot r^3 = 32.
$$[/tex]
Solve for [tex]\( r^3 \)[/tex]:
[tex]$$
r^3 = \frac{32}{3125}.
$$[/tex]
Taking the cube root of both sides, we have
[tex]$$
r = \left(\frac{32}{3125}\right)^{\frac{1}{3}} \approx 0.2171534093.
$$[/tex]

### Step 2. Find the Geometric Means

The first geometric mean [tex]\( GM_1 \)[/tex] is
[tex]$$
GM_1 = ar = 3125 \cdot 0.2171534093 \approx 678.60440415,
$$[/tex]
and the second geometric mean [tex]\( GM_2 \)[/tex] is
[tex]$$
GM_2 = ar^2 = 3125 \cdot (0.2171534093)^2 \approx 147.36125995.
$$[/tex]

### Verification

To verify the progression, note that multiplying [tex]\( GM_2 \)[/tex] by [tex]\( r \)[/tex] should yield the fourth term:
[tex]$$
GM_2 \cdot r = 3125 \cdot r^3 \approx 32.
$$[/tex]

### Final Answer

Thus, the two geometric means that can be inserted between [tex]\(3125\)[/tex] and [tex]\(32\)[/tex] are approximately
[tex]$$
\boxed{678.60 \quad \text{and} \quad 147.36.}
$$[/tex]

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