Let `p=1+1/(sqrt(2))+1/(sqrt(3))+...+1/(sqrt(120))` and `q=1/(sqrt(2))+1/(sqrt(3))+...+1/(sqrt(121))` then

Options

A. `pgt20`
B. `qlt20`
C. `p+qlt40`
D. `p+qgt40`

Explanation

To solve the problem, we need to evaluate the sums \( p \) and \( q \) and use the properties of definite integrals to establish a relationship between them.

### Step 1: Define \( p \) and \( q \)

We have:
\[ p = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{120}} \]
\[ q = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{121}} \]

### Step 2: Express \( p \) in terms of \( q \)

Notice that:
\[ p = 1 + q \]

### Step 3: Set up the definite integral

We know that:
\[ \int_1^{121} \frac{1}{\sqrt{x}} \, dx \]
Calculating this integral:
\[ \int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} \]
Thus,
\[ \int_1^{121} \frac{1}{\sqrt{x}} \, dx = 2\sqrt{121} - 2\sqrt{1} = 2 \cdot 11 - 2 \cdot 1 = 22 - 2 = 20 \]

### Step 4: Establish inequalities

Since \( \frac{1}{\sqrt{x}} \) is a decreasing function, we can use the properties of integrals to establish:
\[ \int_1^{121} \frac{1}{\sqrt{x}} \, dx < p < \int_1^{120} \frac{1}{\sqrt{x}} \, dx \]
and
\[ \int_1^{120} \frac{1}{\sqrt{x}} \, dx < q < \int_1^{121} \frac{1}{\sqrt{x}} \, dx \]

### Step 5: Evaluate the integrals

Calculating \( \int_1^{120} \frac{1}{\sqrt{x}} \, dx \):
\[ \int_1^{120} \frac{1}{\sqrt{x}} \, dx = 2\sqrt{120} - 2\sqrt{1} = 2\sqrt{120} - 2 \]

### Step 6: Compare \( p \) and \( q \)

From the inequalities, we have:
\[ 20 < p < 2\sqrt{120} - 2 \]
and
\[ 20 < q < 20 \]

### Step 7: Combine inequalities

From \( p = 1 + q \), we can substitute:
\[ p + q > 40 \]

### Conclusion

Thus, we conclude that:
\[ p + q > 40 \]

### Final Answer

The answer to the question is \( p + q > 40 \).

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