Explanation:
Okay, so I need to find the eighth term of this geometric sequence: 125/9, 25/3, 5, 3, ... Hmm, let me remember what a geometric sequence is. It's a sequence where each term is multiplied by a common ratio to get the next term. So first, I need to figure out what the common ratio is here.

Let me start by dividing the second term by the first term to see if I can find the ratio. The first term is 125/9 and the second term is 25/3. So, dividing 25/3 by 125/9. Dividing by a fraction is the same as multiplying by its reciprocal, so that would be 25/3 * 9/125. Let me compute that.

First, simplify the numbers. 25 and 125: 25 divides into 125 five times. So 25/125 is 1/5. Then 9/3 is 3. So multiplying those together: 1/5 * 3 = 3/5. Wait, so the ratio is 3/5? Let me check with the next terms to make sure.

The second term is 25/3, and the third term is 5. Let me divide 5 by 25/3. That would be 5 * 3/25 = 15/25 = 3/5. Okay, same ratio. Now check from the third term to the fourth term: 3 divided by 5. Wait, 3 divided by 5 is 3/5. So yes, that's the same ratio. So the common ratio r is 3/5.

Alright, so now that I know the first term a₁ is 125/9 and the common ratio r is 3/5. The formula for the nth term of a geometric sequence is a_n = a₁ * r^(n-1). So, to find the eighth term, I need to plug n = 8 into that formula.

Let me write that out: a₈ = (125/9) * (3/5)^(8-1) = (125/9) * (3/5)^7. Now, I need to compute (3/5)^7 and then multiply it by 125/9.

First, let me compute (3/5)^7. Let me think about how to simplify this. Maybe I can write out the exponents step by step. But that might be tedious. Alternatively, I can simplify the expression before calculating. Let's see.

We have 125 in the numerator and 9 in the denominator. Also, (3/5)^7 is 3^7 / 5^7. So, combining these, the entire expression becomes (125 * 3^7) / (9 * 5^7). Let me see if I can simplify this fraction.

First, note that 125 is 5^3. So replacing that, we have (5^3 * 3^7) / (9 * 5^7). Then, notice that 9 is 3^2. So substituting that, we get (5^3 * 3^7) / (3^2 * 5^7).

Now, let's simplify the exponents. For the 5s: 5^3 divided by 5^7 is 5^(3-7) = 5^(-4) = 1/5^4. For the 3s: 3^7 divided by 3^2 is 3^(7-2) = 3^5.

Therefore, combining these, we have 3^5 / 5^4. Let me compute that.

First, compute 3^5: 3*3=9, 9*3=27, 27*3=81, 81*3=243. So 3^5 is 243.

Then, compute 5^4: 5*5=25, 25*5=125, 125*5=625. So 5^4 is 625.

Therefore, the eighth term is 243/625.

Wait a second, let me double-check my steps to make sure I didn't make a mistake.

Original terms: 125/9, 25/3, 5, 3,...

Common ratio r: 25/3 divided by 125/9 = (25/3)*(9/125) = (25*9)/(3*125). Simplify: 25/125 is 1/5, 9/3 is 3. So 1/5 * 3 = 3/5. Correct.

So the formula: a_n = a₁ * r^(n-1).

For n=8, a₈ = (125/9) * (3/5)^7.

Breaking down: 125/9 * (3^7 / 5^7).

125 is 5³, so 5³ * 3^7 / (9 * 5^7) = 3^7 / (9 * 5^(7-3)) since 5³ cancels with 5^7? Wait, hold on. Wait, original expression:

(5³ * 3^7) / (3² * 5^7) = (3^7 / 3²) * (5³ / 5^7) = 3^(5) * 5^(-4) = 3^5 / 5^4. Yes, that's correct.

3^5 is 243 and 5^4 is 625. Therefore, 243/625.

But let me verify by calculating the terms step by step up to the 8th term, just to make sure.

First term: 125/9 ≈ 13.888...

Second term: 125/9 * 3/5 = (125*3)/(9*5) = (375)/(45) = 25/3 ≈ 8.333... Correct.

Third term: 25/3 * 3/5 = (25/3)*(3/5) = 25/5 = 5. Correct.

Fourth term: 5 * 3/5 = 3. Correct.

Fifth term: 3 * 3/5 = 9/5 = 1.8.

Sixth term: 9/5 * 3/5 = 27/25 = 1.08.

Seventh term: 27/25 * 3/5 = 81/125 = 0.648.

Eighth term: 81/125 * 3/5 = 243/625.

Yes, that's 243 divided by 625. Which is 0.3888...

So that seems consistent. So 243/625 is the eighth term.

Alternatively, if I wanted to write it as a reduced fraction, but 243 and 625: Let's check if they have any common factors.

243 is 3^5. 625 is 5^4. Since they are different primes, they don't share any common factors. Therefore, 243/625 is already in simplest terms.

Therefore, the answer is 243/625.

But let me confirm once again with another approach. Maybe using exponents step by step.

Alternatively, since each term is multiplied by 3/5, starting from 125/9:

Term 1: 125/9

Term 2: 125/9 * 3/5 = (125*3)/(9*5) = (25*3)/9 = 75/9 = 25/3. Correct.

Term 3: 25/3 * 3/5 = 25/5 = 5. Correct.

Term 4: 5 * 3/5 = 3. Correct.

Term 5: 3 * 3/5 = 9/5.

Term 6: 9/5 * 3/5 = 27/25.

Term 7: 27/25 * 3/5 = 81/125.

Term 8: 81/125 * 3/5 = 243/625.

Yes, same result. So that seems to check out.

Alternatively, if I wanted to use the formula for the nth term, which is a₁*r^(n-1).

a₁ is 125/9, r is 3/5, n=8.

So compute (3/5)^7 first. Let me compute that fractionally:

(3/5)^7 = 3^7 / 5^7 = 2187 / 78125.

Then multiply by 125/9:

(125/9) * (2187 / 78125). Let me compute that.

First, note that 125 and 78125: 78125 divided by 125 is 625, since 125*625 = 78125.

Similarly, 2187 divided by 9 is 243. Let me verify that: 9*243 = 2187. Yes, because 200*9=1800, 40*9=360, 3*9=27; 1800+360=2160 +27=2187.

So simplifying the fractions:

(125/9) * (2187 / 78125) = (125 / 78125) * (2187 / 9) = (1/625) * 243 = 243 / 625.

Exactly! So that confirms it again. Therefore, the eighth term is indeed 243/625.

I think that's thorough. I checked multiple ways, step by step, and all lead to the same answer. So I'm confident that 243/625 is the correct eighth term.

**Final Answer**
The eighth term of the sequence is \boxed{\dfrac{243}{625}}.

To find the **eighth term** of the geometric sequence:

$$
\frac{125}{9}, \frac{25}{3}, 5, 3, \ldots
$$

we start by identifying the **common ratio** $ r $. In a geometric sequence, each term is obtained by multiplying the previous term by $ r $, so:

$$
r = \frac{\text{second term}}{\text{first term}} = \frac{\frac{25}{3}}{\frac{125}{9}} = \frac{25}{3} \times \frac{9}{125} = \frac{225}{375} = \frac{3}{5}
$$

We verify this with the next term:

$$
\frac{25}{3} \times \frac{3}{5} = 5, \quad \text{which matches the third term.}
$$

So, the common ratio is $ r = \frac{3}{5} $, and the first term is $ a = \frac{125}{9} $.

---

The general formula for the $ n $-th term of a geometric sequence is:

$$
a_n = a \cdot r^{n-1}
$$

Substituting $ a = \frac{125}{9} $, $ r = \frac{3}{5} $, and $ n = 8 $:

$$
a_8 = \frac{125}{9} \cdot \left(\frac{3}{5}\right)^7
$$

We simplify $ \left(\frac{3}{5}\right)^7 $:

$$
\left(\frac{3}{5}\right)^7 = \frac{3^7}{5^7} = \frac{2187}{78125}
$$

Now multiply:

$$
a_8 = \frac{125}{9} \cdot \frac{2187}{78125}
$$

Simplify:

- $ \frac{125}{78125} = \frac{1}{625} $
- $ \frac{2187}{9} = 243 $

So:

$$
a_8 = \frac{1}{625} \cdot 243 = \frac{243}{625}
$$

---

**Final Answer**:
$$
\boxed{\dfrac{243}{625}}
$$

[Assistant]
Explanation:
Okay, so I need to find the eighth term of this geometric sequence: 125/9, 25/3, 5, 3, ... Hmm, let me remember what a geometric sequence is. It's a sequence where each term is multiplied by a common ratio to get the next term. So first, I need to figure out what the common ratio is here.

Let me start by dividing the second term by the first term to see if I can find the ratio. The first term is 125/9 and the second term is 25/3. So, dividing 25/3 by 125/9. Dividing by a fraction is the same as multiplying by its reciprocal, so that would be 25/3 * 9/125. Let me compute that.

First, simplify the numbers. 25 and 125: 25 divides into 125 five times. So 25/125 is 1/5. Then 9/3 is 3. So multiplying those together: 1/5 * 3 = 3/5. Wait, so the ratio is 3/5? Let me check with the next terms to make sure.

The second term is 25/3, and the third term is 5. Let me divide 5 by 25/3. That would be 5 * 3/25 = 15/25 = 3/5. Okay, same ratio. Now check from the third term to the fourth term: 3 divided by 5. Wait, 3 divided by 5 is 3/5. So yes, that's the same ratio. So the common ratio r is 3/5.

Alright, so now that I know the first term a₁ is 125/9 and the common ratio r is 3/5. The formula for the nth term of a geometric sequence is a_n = a₁ * r^(n-1). So, to find the eighth term, I need to plug n = 8 into that formula.

Let me write that out: a₈ = (125/9) * (3/5)^(8-1) = (125/9) * (3/5)^7. Now, I need to compute (3/5)^7 and then multiply it by 125/9.

First, let me compute (3/5)^7. Let me think about how to simplify this. Maybe I can write out the exponents step by step. But that might be tedious. Alternatively, I can simplify the expression before calculating. Let's see.

We have 125 in the numerator and 9 in the denominator. Also, (3/5)^7 is 3^7 / 5^7. So, combining these, the entire expression becomes (125 * 3^7) / (9 * 5^7). Let me see if I can simplify this fraction.

First, note that 125 is 5^3. So replacing that, we have (5^3 * 3^7) / (9 * 5^7). Then, notice that 9 is 3^2. So substituting that, we get (5^3 * 3^7) / (3^2 * 5^7).

Now, let's simplify the exponents. For the 5s: 5^3 divided by 5^7 is 5^(3-7) = 5^(-4) = 1/5^4. For the 3s: 3^7 divided by 3^2 is 3^(7-2) = 3^5.

Therefore, combining these, we have 3^5 / 5^4. Let me compute that.

First, compute 3^5: 3*3=9, 9*3=27, 27*3=81, 81*3=243. So 3^5 is 243.

Then, compute 5^4: 5*5=25, 25*5=125, 125*5=625. So 5^4 is 625.

Therefore, the eighth term is 243/625.

Wait a second, let me double-check my steps to make sure I didn't make a mistake.

Original terms: 125/9, 25/3, 5, 3,...

Common ratio r: 25/3 divided by 125/9 = (25/3)*(9/125) = (25*9)/(3*125). Simplify: 25/125 is 1/5, 9/3 is 3. So 1/5 * 3 = 3/5. Correct.

So the formula: a_n = a₁ * r^(n-1).

For n=8, a₈ = (125/9) * (3/5)^7.

Breaking down: 125/9 * (3^7 / 5^7).

125 is 5³, so 5³ * 3^7 / (9 * 5^7) = 3^7 / (9 * 5^(7-3)) since 5³ cancels with 5^7? Wait, hold on. Wait, original expression:

(5³ * 3^7) / (3² * 5^7) = (3^7 / 3²) * (5³ / 5^7) = 3^(5) * 5^(-4) = 3^5 / 5^4. Yes, that's correct.

3^5 is 243 and 5^4 is 625. Therefore, 243/625.

But let me verify by calculating the terms step by step up to the 8th term, just to make sure.

First term: 125/9 ≈ 13.888...

Second term: 125/9 * 3/5 = (125*3)/(9*5) = (375)/(45) = 25/3 ≈ 8.333... Correct.

Third term: 25/3 * 3/5 = (25/3)*(3/5) = 25/5 = 5. Correct.

Fourth term: 5 * 3/5 = 3. Correct.

Fifth term: 3 * 3/5 = 9/5 = 1.8.

Sixth term: 9/5 * 3/5 = 27/25 = 1.08.

Seventh term: 27/25 * 3/5 = 81/125 = 0.648.

Eighth term: 81/125 * 3/5 = 243/625.

Yes, that's 243 divided by 625. Which is 0.3888...

So that seems consistent. So 243/625 is the eighth term.

Alternatively, if I wanted to write it as a reduced fraction, but 243 and 625: Let's check if they have any common factors.

243 is 3^5. 625 is 5^4. Since they are different primes, they don't share any common factors. Therefore, 243/625 is already in simplest terms.

Therefore, the answer is 243/625.

But let me confirm once again with another approach. Maybe using exponents step by step.

Alternatively, since each term is multiplied by 3/5, starting from 125/9:

Term 1: 125/9

Term 2: 125/9 * 3/5 = (125*3)/(9*5) = (25*3)/9 = 75/9 = 25/3. Correct.

Term 3: 25/3 * 3/5 = 25/5 = 5. Correct.

Term 4: 5 * 3/5 = 3. Correct.

Term 5: 3 * 3/5 = 9/5.

Term 6: 9/5 * 3/5 = 27/25.

Term 7: 27/25 * 3/5 = 81/125.

Term 8: 81/125 * 3/5 = 243/625.

Yes, same result. So that seems to check out.

Alternatively, if I wanted to use the formula for the nth term, which is a₁*r^(n-1).

a₁ is 125/9, r is 3/5, n=8.

So compute (3/5)^7 first. Let me compute that fractionally:

(3/5)^7 = 3^7 / 5^7 = 2187 / 78125.

Then multiply by 125/9:

(125/9) * (2187 / 78125). Let me compute that.

First, note that 125 and 78125: 78125 divided by 125 is 625, since 125*625 = 78125.

Similarly, 2187 divided by 9 is 243. Let me verify that: 9*243 = 2187. Yes, because 200*9=1800, 40*9=360, 3*9=27; 1800+360=2160 +27=2187.

So simplifying the fractions:

(125/9) * (2187 / 78125) = (125 / 78125) * (2187 / 9) = (1/625) * 243 = 243 / 625.

Exactly! So that confirms it again. Therefore, the eighth term is indeed 243/625.

I think that's thorough. I checked multiple ways, step by step, and all lead to the same answer. So I'm confident that 243/625 is the correct eighth term.

**Final Answer**
The eighth term of the sequence is \boxed{\dfrac{243}{625}}.

To find the **eighth term** of the geometric sequence:

$$
\frac{125}{9}, \frac{25}{3}, 5, 3, \ldots
$$

we start by identifying the **common ratio** $ r $. In a geometric sequence, each term is obtained by multiplying the previous term by $ r $, so:

$$
r = \frac{\text{second term}}{\text{first term}} = \frac{\frac{25}{3}}{\frac{125}{9}} = \frac{25}{3} \times \frac{9}{125} = \frac{225}{375} = \frac{3}{5}
$$

We verify this with the next term:

$$
\frac{25}{3} \times \frac{3}{5} = 5, \quad \text{which matches the third term.}
$$

So, the common ratio is $ r = \frac{3}{5} $, and the first term is $ a = \frac{125}{9} $.

---

The general formula for the $ n $-th term of a geometric sequence is:

$$
a_n = a \cdot r^{n-1}
$$

Substituting $ a = \frac{125}{9} $, $ r = \frac{3}{5} $, and $ n = 8 $:

$$
a_8 = \frac{125}{9} \cdot \left(\frac{3}{5}\right)^7
$$

We simplify $ \left(\frac{3}{5}\right)^7 $:

$$
\left(\frac{3}{5}\right)^7 = \frac{3^7}{5^7} = \frac{2187}{78125}
$$

Now multiply:

$$
a_8 = \frac{125}{9} \cdot \frac{2187}{78125}
$$

Simplify:

- $ \frac{125}{78125} = \frac{1}{625} $
- $ \frac{2187}{9} = 243 $

So:

$$
a_8 = \frac{1}{625} \cdot 243 = \frac{243}{625}
$$

---

**Final Answer**:
$$
\boxed{\dfrac{243}{625}}
$$

Solution:
The common ratio between consecutive terms is $\frac{3}{5}$ (you can choose any two consecutive terms and divide the second one by the first to find the common ratio). So the $n^\text{th}$ term of the sequence is $\frac{125}{9} \cdot \left( \frac{3}{5} \right)^{n-1}$. Plugging in $n=8$, we get $$
\frac{125}{9} \cdot \left( \frac{3}{5} \right)^{7} = \frac{5^3}{3^2} \cdot \frac{3^7}{5^7}
= \frac{3^5}{5^4}
= \boxed{\frac{243}{625}}.
$$


Explanation:
Okay, so I need to find the constant term in the expansion of $\left(10x^3 - \frac{1}{2x^2}\right)^5$. Hmm, let me think about how to approach this.

First, I remember that when you have a binomial raised to a power, like $(a + b)^n$, you can use the binomial theorem to expand it. The binomial theorem says that each term in the expansion is of the form $\binom{n}{k} a^{n-k} b^k$, where $k$ ranges from 0 to $n$.

In this case, our binomial is $\left(10x^3 - \frac{1}{2x^2}\right)^5$, so $a = 10x^3$, $b = -\frac{1}{2x^2}$, and $n = 5$.

So, the general term in the expansion would be $\binom{5}{k} (10x^3)^{5 - k} \left(-\frac{1}{2x^2}\right)^k$.

Now, I need to find the term that is a constant term. That means the exponent of $x$ in that term should be zero. So, I need to figure out which value of $k$ makes the exponent of $x$ equal to zero.

Let me first figure out the exponent of $x$ in the general term. Let's break it down:

The first part is $(10x^3)^{5 - k}$, which would contribute $x^{3(5 - k)}$.

The second part is $\left(-\frac{1}{2x^2}\right)^k$, which is the same as $(-1)^k \cdot \frac{1}{2^k} \cdot x^{-2k}$.

So when we multiply these two parts together, the exponent of $x$ would be $3(5 - k) - 2k$. Let me compute that:

$3(5 - k) - 2k = 15 - 3k - 2k = 15 - 5k$.

We want this exponent to be zero, so set up the equation:

$15 - 5k = 0$.

Solving for $k$:

$15 = 5k$

$k = 3$.

Okay, so the constant term is when $k = 3$. Now, let me compute that term.

First, compute the binomial coefficient $\binom{5}{3}$.

$\binom{5}{3} = \frac{5!}{3! \cdot (5 - 3)!} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = \frac{120}{12} = 10$.

Next, compute $(10x^3)^{5 - 3}$. Since $5 - 3 = 2$, that's $(10x^3)^2 = 10^2 \cdot x^{6} = 100x^6$.

Then, compute $\left(-\frac{1}{2x^2}\right)^3$. Let's break that down:

First, $-\frac{1}{2x^2}$ raised to the 3rd power is $(-1)^3 \cdot \frac{1}{2^3} \cdot x^{-2 \cdot 3} = -1 \cdot \frac{1}{8} \cdot x^{-6} = -\frac{1}{8}x^{-6}$.

Now, multiply those two parts together:

$100x^6 \cdot \left(-\frac{1}{8}x^{-6}\right)$.

Multiplying the coefficients: $100 \times -\frac{1}{8} = -\frac{100}{8} = -12.5$.

Multiplying the $x$ terms: $x^6 \times x^{-6} = x^{0} = 1$.

So, the term simplifies to $-12.5 \times 1 = -12.5$.

But wait, the question says "Find the constant term," and they might expect an exact fraction instead of a decimal. Let me check my calculation again.

Original coefficient: $\binom{5}{3} = 10$, then $(10x^3)^2 = 100x^6$, and $\left(-\frac{1}{2x^2}\right)^3 = -\frac{1}{8x^6}$.

Multiplying all together: $10 \times 100 \times \left(-\frac{1}{8}\right)$ for the coefficients, and $x^6 \times x^{-6} = 1$.

Wait a second, I think I made a mistake earlier when I multiplied the coefficients. Let me redo that part.

The binomial coefficient is 10, then $(10x^3)^{2}$ gives 100, and $\left(-\frac{1}{2x^2}\right)^3$ gives $-1/8$. So multiplying them all together: 10 * 100 * (-1/8).

Let me compute that:

First, 10 * 100 = 1000.

Then, 1000 * (-1/8) = -1000/8 = -125/1 = -125? Wait, 1000 divided by 8 is 125? Wait 8*125 is 1000, yes. So that would be -125.

Wait, hold on, but earlier I thought 100 * (-1/8) is -12.5, but I forgot that the binomial coefficient is 10. So actually, the coefficient is 10 * 100 * (-1/8) = -1000/8 = -125. So that's -125.

But earlier, I had said -12.5, which was wrong because I didn't multiply by the binomial coefficient. So my mistake was there. Let me make sure.

Yes, the general term is $\binom{5}{k} \times (10x^3)^{5 - k} \times \left(-\frac{1}{2x^2}\right)^k$. So when k=3:

$\binom{5}{3} = 10$

$(10x^3)^{2} = 100x^6$

$\left(-\frac{1}{2x^2}\right)^3 = -\frac{1}{8x^6}$

Multiplying all together: 10 * 100x^6 * (-1/8x^6) = 10 * 100 * (-1/8) * x^6/x^6 = 10 * (-100/8) * 1 = 10 * (-12.5) = -125.

Yes, that's right. So the constant term is -125.

But let me check again step by step to make sure. Maybe I can compute the coefficient in another way.

Let me write the entire term:

$\binom{5}{3} \times (10)^{2} \times (-1/2)^3 \times x^{3(5 - 3)} \times x^{-2(3)}$.

Calculating the coefficients:

$\binom{5}{3} = 10$

$(10)^{2} = 100$

$(-1/2)^3 = -1/8$

Multiply them all: 10 * 100 * (-1/8) = 10 * (-100/8) = 10 * (-25/2) = -250/2 = -125.

Yes, same result. So the coefficient is -125, and the x terms cancel out, so the constant term is -125.

Wait, but let me check if there is another term that could contribute. For example, maybe when k = something else? Let me verify if k=3 is the only possible value where the exponent is zero.

Earlier, I had the exponent as 15 - 5k. So setting 15 - 5k = 0 gives k=3, which is an integer between 0 and 5. So that's the only term. Therefore, the constant term is indeed -125.

But just to be thorough, let me check for k=0,1,2,3,4,5 what the exponent is.

For k=0: exponent = 15 - 0 = 15. Not zero.

k=1: 15 - 5 = 10. Not zero.

k=2: 15 -10=5. Not zero.

k=3: 15 -15=0. Okay.

k=4: 15 -20= -5. Not zero.

k=5: 15 -25= -10. Not zero.

So yes, only when k=3. So that term is the only constant term.

Therefore, the constant term is -125.

But wait, the problem says "the constant term in the expansion". So, the answer should be -125? Let me check again.

Alternatively, maybe I made a mistake in the sign? Let me check the sign again.

The term is $\binom{5}{3} \times (10x^3)^{2} \times (-1/2x^2)^3$.

Breaking down the sign:

The second term is positive because (10x^3)^{2} is positive. The third term is (-1/2x^2)^3, which is negative because it's raised to an odd power. So the sign is negative.

So overall, the coefficient is negative. So -125 is correct.

Alternatively, maybe I should write it as a fraction? -125 is an integer, so that's fine.

But let me check once more by expanding the expression step by step.

Alternatively, let me think of the expansion as:

Each term is C(5, k) * (10x^3)^{5 - k} * (-1/(2x^2))^k.

So, for k=3:

C(5,3) is 10. Then 10x^3 to the (5 - 3) = 2: 10^2 x^6 = 100x^6. Then (-1/(2x^2))^3 is (-1)^3 / (2^3 x^6) = -1/(8x^6).

Multiplying together: 10 * 100 * (-1/8) * x^6 / x^6 = 10 * 100 * (-1/8) * 1.

Which is 10 * (-100/8) = 10 * (-25/2) = -250/2 = -125. Yep, same result.

So I think that's correct.

Wait, but let me check a different approach. Maybe plug in x=1 and see what the expansion evaluates to? Although that might not directly give the constant term, but maybe as a check?

Wait, if I plug in x=1, the entire expression becomes (10*1^3 - 1/(2*1^2))^5 = (10 - 1/2)^5 = (19/2)^5.

But the expansion evaluated at x=1 would be sum_{k=0}^5 [C(5, k) * (10)^{5 -k} * (-1/2)^k].

But the constant term is -125, so if we sum all the terms, when x=1, the value is (19/2)^5. Let's compute that.

But maybe that's too time-consuming. Alternatively, perhaps check if the constant term is indeed -125.

Alternatively, think about whether the answer makes sense.

Let me compute the value of the term for k=3.

Yes, as above, that term is -125.

Alternatively, think about another term. For example, if I take k=1, what's the term?

k=1: exponent is 15 -5*1=10. So term is C(5,1)*(10x^3)^4*(-1/(2x^2))^1 = 5*10^4 x^{12} * (-1/(2x^2)) = 5*10000*(-1/2) x^{10} = 5* (-5000) x^{10} = -25000x^{10}.

Similarly, for k=2: exponent is 5, term would be C(5,2)*(10x^3)^3*(-1/(2x^2))^2 = 10 * 1000x^9 * (1/(4x^4)) = 10 * 1000/4 x^{5} = 10 * 250 x^5 = 2500x^5.

So those terms have exponents 10, 5, etc. So for k=3, exponent is zero, so the term is -125.

For k=4, exponent is -5, term would be C(5,4)*(10x^3)^1*(-1/(2x^2))^4 = 5 * 10x^3 * (1/(16x^8)) = 5*10/(16) x^{-5} = 50/16 x^{-5} = 25/8 x^{-5}.

k=5: exponent is -10, term is C(5,5)*(10x^3)^0*(-1/(2x^2))^5 = 1 * 1 * (-1)^5/(32x^{10}) = -1/(32x^{10}).

Therefore, the expansion includes terms with exponents 15, 10, 5, 0, -5, -10. So the constant term is indeed when exponent is 0, which is k=3, and that term is -125.

Therefore, I think my answer is correct.

So the constant term is -125.

But wait, the problem says "the constant term", so maybe they want it in boxed form. Let me check once again for any miscalculations.

Another way: Let me think of the general term again.

Term is $\binom{5}{k} \times (10)^{5 - k} \times (-1/2)^k \times x^{3(5 - k) - 2k}$.

Which simplifies to $\binom{5}{k} \times 10^{5 -k} \times (-1)^k \times 2^{-k} \times x^{15 -5k}$.

So the exponent is 15 -5k. So setting 15 -5k =0 gives k=3.

Then the coefficient is $\binom{5}{3} \times 10^{2} \times (-1)^3 \times 2^{-3}$.

Which is 10 * 100 * (-1) * (1/8).

Which is 10 * 100 * (-1)/8 = 10 * (-100)/8 = 10 * (-25)/2 = -250 / 2 = -125.

Yep, same answer.

So I think I can be confident that the constant term is -125.

But maybe I need to check if the original expression was written correctly?

Original problem: $\left(10x^3 - \frac{1}{2x^2}\right)^5$. Yes, that's what I used.

So the expansion of that is correct.

Therefore, the constant term is -125.

**Final Answer**
The constant term is \boxed{-125}.

To find the **constant term** in the expansion of the expression:

$$
\left(10x^3 - \frac{1}{2x^2}\right)^5
$$

we use the **Binomial Theorem**, which tells us that the general term in the expansion of $(a + b)^n$ is:

$$
T_k = \binom{n}{k} a^{n-k} b^k
$$

---

### Step 1: Identify $a$, $b$, and $n$

In our case:

- $a = 10x^3$
- $b = -\frac{1}{2x^2}$
- $n = 5$

So, the general term becomes:

$$
T_k = \binom{5}{k} (10x^3)^{5-k} \left(-\frac{1}{2x^2}\right)^k
$$

---

### Step 2: Determine the exponent of $x$ in a general term

Break it down:

- $(10x^3)^{5-k} = 10^{5-k} x^{3(5-k)}$
- $\left(-\frac{1}{2x^2}\right)^k = \left(-1\right)^k \cdot \left(\frac{1}{2}\right)^k \cdot x^{-2k}$

Multiply them:

$$
T_k = \binom{5}{k} \cdot 10^{5-k} \cdot \left(-\frac{1}{2}\right)^k \cdot x^{3(5-k) - 2k}
$$

Simplify the exponent of $x$:

$$
3(5 - k) - 2k = 15 - 3k - 2k = 15 - 5k
$$

So, the exponent of $x$ in the general term is $15 - 5k$. To find the **constant term**, we set this equal to zero:

$$
15 - 5k = 0 \Rightarrow k = 3
$$

---

### Step 3: Compute the constant term when $k = 3$

Now plug $k = 3$ into the general term expression:

$$
T_3 = \binom{5}{3} \cdot (10x^3)^{5 - 3} \cdot \left(-\frac{1}{2x^2}\right)^3
$$

Compute each part:

- $\binom{5}{3} = 10$
- $(10x^3)^2 = 10^2 x^6 = 100x^6$
- $\left(-\frac{1}{2x^2}\right)^3 = (-1)^3 \cdot \frac{1}{2^3} \cdot x^{-6} = -\frac{1}{8x^6}$

Now multiply all together:

$$
T_3 = 10 \cdot 100x^6 \cdot \left(-\frac{1}{8x^6}\right)
$$

$$
T_3 = 10 \cdot 100 \cdot \left(-\frac{1}{8}\right) \cdot \frac{x^6}{x^6} = 10 \cdot \left(-\frac{100}{8}\right) = \frac{-1000}{8} = -125
$$

---

### Step 4: Final Answer

Thus, the **constant term** in the expansion is:

$$
\boxed{-125}
$$

[Assistant]
Explanation:
Okay, so I need to find the constant term in the expansion of $\left(10x^3 - \frac{1}{2x^2}\right)^5$. Hmm, let me think about how to approach this.

First, I remember that when you have a binomial raised to a power, like $(a + b)^n$, you can use the binomial theorem to expand it. The binomial theorem says that each term in the expansion is of the form $\binom{n}{k} a^{n-k} b^k$, where $k$ ranges from 0 to $n$.

In this case, our binomial is $\left(10x^3 - \frac{1}{2x^2}\right)^5$, so $a = 10x^3$, $b = -\frac{1}{2x^2}$, and $n = 5$.

So, the general term in the expansion would be $\binom{5}{k} (10x^3)^{5 - k} \left(-\frac{1}{2x^2}\right)^k$.

Now, I need to find the term that is a constant term. That means the exponent of $x$ in that term should be zero. So, I need to figure out which value of $k$ makes the exponent of $x$ equal to zero.

Let me first figure out the exponent of $x$ in the general term. Let's break it down:

The first part is $(10x^3)^{5 - k}$, which would contribute $x^{3(5 - k)}$.

The second part is $\left(-\frac{1}{2x^2}\right)^k$, which is the same as $(-1)^k \cdot \frac{1}{2^k} \cdot x^{-2k}$.

So when we multiply these two parts together, the exponent of $x$ would be $3(5 - k) - 2k$. Let me compute that:

$3(5 - k) - 2k = 15 - 3k - 2k = 15 - 5k$.

We want this exponent to be zero, so set up the equation:

$15 - 5k = 0$.

Solving for $k$:

$15 = 5k$

$k = 3$.

Okay, so the constant term is when $k = 3$. Now, let me compute that term.

First, compute the binomial coefficient $\binom{5}{3}$.

$\binom{5}{3} = \frac{5!}{3! \cdot (5 - 3)!} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = \frac{120}{12} = 10$.

Next, compute $(10x^3)^{5 - 3}$. Since $5 - 3 = 2$, that's $(10x^3)^2 = 10^2 \cdot x^{6} = 100x^6$.

Then, compute $\left(-\frac{1}{2x^2}\right)^3$. Let's break that down:

First, $-\frac{1}{2x^2}$ raised to the 3rd power is $(-1)^3 \cdot \frac{1}{2^3} \cdot x^{-2 \cdot 3} = -1 \cdot \frac{1}{8} \cdot x^{-6} = -\frac{1}{8}x^{-6}$.

Now, multiply those two parts together:

$100x^6 \cdot \left(-\frac{1}{8}x^{-6}\right)$.

Multiplying the coefficients: $100 \times -\frac{1}{8} = -\frac{100}{8} = -12.5$.

Multiplying the $x$ terms: $x^6 \times x^{-6} = x^{0} = 1$.

So, the term simplifies to $-12.5 \times 1 = -12.5$.

But wait, the question says "Find the constant term," and they might expect an exact fraction instead of a decimal. Let me check my calculation again.

Original coefficient: $\binom{5}{3} = 10$, then $(10x^3)^2 = 100x^6$, and $\left(-\frac{1}{2x^2}\right)^3 = -\frac{1}{8x^6}$.

Multiplying all together: $10 \times 100 \times \left(-\frac{1}{8}\right)$ for the coefficients, and $x^6 \times x^{-6} = 1$.

Wait a second, I think I made a mistake earlier when I multiplied the coefficients. Let me redo that part.

The binomial coefficient is 10, then $(10x^3)^{2}$ gives 100, and $\left(-\frac{1}{2x^2}\right)^3$ gives $-1/8$. So multiplying them all together: 10 * 100 * (-1/8).

Let me compute that:

First, 10 * 100 = 1000.

Then, 1000 * (-1/8) = -1000/8 = -125/1 = -125? Wait, 1000 divided by 8 is 125? Wait 8*125 is 1000, yes. So that would be -125.

Wait, hold on, but earlier I thought 100 * (-1/8) is -12.5, but I forgot that the binomial coefficient is 10. So actually, the coefficient is 10 * 100 * (-1/8) = -1000/8 = -125. So that's -125.

But earlier, I had said -12.5, which was wrong because I didn't multiply by the binomial coefficient. So my mistake was there. Let me make sure.

Yes, the general term is $\binom{5}{k} \times (10x^3)^{5 - k} \times \left(-\frac{1}{2x^2}\right)^k$. So when k=3:

$\binom{5}{3} = 10$

$(10x^3)^{2} = 100x^6$

$\left(-\frac{1}{2x^2}\right)^3 = -\frac{1}{8x^6}$

Multiplying all together: 10 * 100x^6 * (-1/8x^6) = 10 * 100 * (-1/8) * x^6/x^6 = 10 * (-100/8) * 1 = 10 * (-12.5) = -125.

Yes, that's right. So the constant term is -125.

But let me check again step by step to make sure. Maybe I can compute the coefficient in another way.

Let me write the entire term:

$\binom{5}{3} \times (10)^{2} \times (-1/2)^3 \times x^{3(5 - 3)} \times x^{-2(3)}$.

Calculating the coefficients:

$\binom{5}{3} = 10$

$(10)^{2} = 100$

$(-1/2)^3 = -1/8$

Multiply them all: 10 * 100 * (-1/8) = 10 * (-100/8) = 10 * (-25/2) = -250/2 = -125.

Yes, same result. So the coefficient is -125, and the x terms cancel out, so the constant term is -125.

Wait, but let me check if there is another term that could contribute. For example, maybe when k = something else? Let me verify if k=3 is the only possible value where the exponent is zero.

Earlier, I had the exponent as 15 - 5k. So setting 15 - 5k = 0 gives k=3, which is an integer between 0 and 5. So that's the only term. Therefore, the constant term is indeed -125.

But just to be thorough, let me check for k=0,1,2,3,4,5 what the exponent is.

For k=0: exponent = 15 - 0 = 15. Not zero.

k=1: 15 - 5 = 10. Not zero.

k=2: 15 -10=5. Not zero.

k=3: 15 -15=0. Okay.

k=4: 15 -20= -5. Not zero.

k=5: 15 -25= -10. Not zero.

So yes, only when k=3. So that term is the only constant term.

Therefore, the constant term is -125.

But wait, the problem says "the constant term in the expansion". So, the answer should be -125? Let me check again.

Alternatively, maybe I made a mistake in the sign? Let me check the sign again.

The term is $\binom{5}{3} \times (10x^3)^{2} \times (-1/2x^2)^3$.

Breaking down the sign:

The second term is positive because (10x^3)^{2} is positive. The third term is (-1/2x^2)^3, which is negative because it's raised to an odd power. So the sign is negative.

So overall, the coefficient is negative. So -125 is correct.

Alternatively, maybe I should write it as a fraction? -125 is an integer, so that's fine.

But let me check once more by expanding the expression step by step.

Alternatively, let me think of the expansion as:

Each term is C(5, k) * (10x^3)^{5 - k} * (-1/(2x^2))^k.

So, for k=3:

C(5,3) is 10. Then 10x^3 to the (5 - 3) = 2: 10^2 x^6 = 100x^6. Then (-1/(2x^2))^3 is (-1)^3 / (2^3 x^6) = -1/(8x^6).

Multiplying together: 10 * 100 * (-1/8) * x^6 / x^6 = 10 * 100 * (-1/8) * 1.

Which is 10 * (-100/8) = 10 * (-25/2) = -250/2 = -125. Yep, same result.

So I think that's correct.

Wait, but let me check a different approach. Maybe plug in x=1 and see what the expansion evaluates to? Although that might not directly give the constant term, but maybe as a check?

Wait, if I plug in x=1, the entire expression becomes (10*1^3 - 1/(2*1^2))^5 = (10 - 1/2)^5 = (19/2)^5.

But the expansion evaluated at x=1 would be sum_{k=0}^5 [C(5, k) * (10)^{5 -k} * (-1/2)^k].

But the constant term is -125, so if we sum all the terms, when x=1, the value is (19/2)^5. Let's compute that.

But maybe that's too time-consuming. Alternatively, perhaps check if the constant term is indeed -125.

Alternatively, think about whether the answer makes sense.

Let me compute the value of the term for k=3.

Yes, as above, that term is -125.

Alternatively, think about another term. For example, if I take k=1, what's the term?

k=1: exponent is 15 -5*1=10. So term is C(5,1)*(10x^3)^4*(-1/(2x^2))^1 = 5*10^4 x^{12} * (-1/(2x^2)) = 5*10000*(-1/2) x^{10} = 5* (-5000) x^{10} = -25000x^{10}.

Similarly, for k=2: exponent is 5, term would be C(5,2)*(10x^3)^3*(-1/(2x^2))^2 = 10 * 1000x^9 * (1/(4x^4)) = 10 * 1000/4 x^{5} = 10 * 250 x^5 = 2500x^5.

So those terms have exponents 10, 5, etc. So for k=3, exponent is zero, so the term is -125.

For k=4, exponent is -5, term would be C(5,4)*(10x^3)^1*(-1/(2x^2))^4 = 5 * 10x^3 * (1/(16x^8)) = 5*10/(16) x^{-5} = 50/16 x^{-5} = 25/8 x^{-5}.

k=5: exponent is -10, term is C(5,5)*(10x^3)^0*(-1/(2x^2))^5 = 1 * 1 * (-1)^5/(32x^{10}) = -1/(32x^{10}).

Therefore, the expansion includes terms with exponents 15, 10, 5, 0, -5, -10. So the constant term is indeed when exponent is 0, which is k=3, and that term is -125.

Therefore, I think my answer is correct.

So the constant term is -125.

But wait, the problem says "the constant term", so maybe they want it in boxed form. Let me check once again for any miscalculations.

Another way: Let me think of the general term again.

Term is $\binom{5}{k} \times (10)^{5 - k} \times (-1/2)^k \times x^{3(5 - k) - 2k}$.

Which simplifies to $\binom{5}{k} \times 10^{5 -k} \times (-1)^k \times 2^{-k} \times x^{15 -5k}$.

So the exponent is 15 -5k. So setting 15 -5k =0 gives k=3.

Then the coefficient is $\binom{5}{3} \times 10^{2} \times (-1)^3 \times 2^{-3}$.

Which is 10 * 100 * (-1) * (1/8).

Which is 10 * 100 * (-1)/8 = 10 * (-100)/8 = 10 * (-25)/2 = -250 / 2 = -125.

Yep, same answer.

So I think I can be confident that the constant term is -125.

But maybe I need to check if the original expression was written correctly?

Original problem: $\left(10x^3 - \frac{1}{2x^2}\right)^5$. Yes, that's what I used.

So the expansion of that is correct.

Therefore, the constant term is -125.

**Final Answer**
The constant term is \boxed{-125}.

To find the **constant term** in the expansion of the expression:

$$
\left(10x^3 - \frac{1}{2x^2}\right)^5
$$

we use the **Binomial Theorem**, which tells us that the general term in the expansion of $(a + b)^n$ is:

$$
T_k = \binom{n}{k} a^{n-k} b^k
$$

---

### Step 1: Identify $a$, $b$, and $n$

In our case:

- $a = 10x^3$
- $b = -\frac{1}{2x^2}$
- $n = 5$

So, the general term becomes:

$$
T_k = \binom{5}{k} (10x^3)^{5-k} \left(-\frac{1}{2x^2}\right)^k
$$

---

### Step 2: Determine the exponent of $x$ in a general term

Break it down:

- $(10x^3)^{5-k} = 10^{5-k} x^{3(5-k)}$
- $\left(-\frac{1}{2x^2}\right)^k = \left(-1\right)^k \cdot \left(\frac{1}{2}\right)^k \cdot x^{-2k}$

Multiply them:

$$
T_k = \binom{5}{k} \cdot 10^{5-k} \cdot \left(-\frac{1}{2}\right)^k \cdot x^{3(5-k) - 2k}
$$

Simplify the exponent of $x$:

$$
3(5 - k) - 2k = 15 - 3k - 2k = 15 - 5k
$$

So, the exponent of $x$ in the general term is $15 - 5k$. To find the **constant term**, we set this equal to zero:

$$
15 - 5k = 0 \Rightarrow k = 3
$$

---

### Step 3: Compute the constant term when $k = 3$

Now plug $k = 3$ into the general term expression:

$$
T_3 = \binom{5}{3} \cdot (10x^3)^{5 - 3} \cdot \left(-\frac{1}{2x^2}\right)^3
$$

Compute each part:

- $\binom{5}{3} = 10$
- $(10x^3)^2 = 10^2 x^6 = 100x^6$
- $\left(-\frac{1}{2x^2}\right)^3 = (-1)^3 \cdot \frac{1}{2^3} \cdot x^{-6} = -\frac{1}{8x^6}$

Now multiply all together:

$$
T_3 = 10 \cdot 100x^6 \cdot \left(-\frac{1}{8x^6}\right)
$$

$$
T_3 = 10 \cdot 100 \cdot \left(-\frac{1}{8}\right) \cdot \frac{x^6}{x^6} = 10 \cdot \left(-\frac{100}{8}\right) = \frac{-1000}{8} = -125
$$

---

### Step 4: Final Answer

Thus, the **constant term** in the expansion is:

$$
\boxed{-125}
$$

Solution: To get a constant term, the exponents of $x$ must cancel.  If we take the term with 2 $x^3$'s and 3 $\frac{1}{x^2}$'s, then they will cancel.  By the Binomial Theorem, this term is $$\binom52 (10x^3)^2\left(-\frac{1}{2x^2}\right)^3=10\cdot100\cdot-\frac{1}{8}\cdot x^6\cdot\frac{1}{x^6}$$$$\Rightarrow \frac{1000}{-8}=\boxed{-125}$$