Finding the Sum of 20 Terms in an AP

Let's break down this problem about Arithmetic Progressions (AP). We are given that the sum of the first 9 terms of an AP is equal to the sum of the first 11 terms. Our goal is to find the sum of the first 20 terms of this AP.

The formula for the sum of the first \(n\) terms of an Arithmetic Progression (AP) is given by:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

Where:

• \(S_n\) is the sum of the first \(n\) terms.
• \(a\) is the first term of the AP.
• \(d\) is the common difference of the AP.
• \(n\) is the number of terms.

We are given that the sum of the first 9 terms (\(S_9\)) is equal to the sum of the first 11 terms (\(S_{11}\)).

Using the formula, we can write:

$$S_9 = \frac{9}{2}[2a + (9-1)d] = \frac{9}{2}[2a + 8d]$$

$$S_{11} = \frac{11}{2}[2a + (11-1)d] = \frac{11}{2}[2a + 10d]$$

According to the problem, \(S_9 = S_{11}\). So, we set the expressions equal to each other:

$$\frac{9}{2}[2a + 8d] = \frac{11}{2}[2a + 10d]$$

We can multiply both sides by 2 to eliminate the denominators:

$$9(2a + 8d) = 11(2a + 10d)$$

Now, let's distribute the numbers on both sides:

$$18a + 72d = 22a + 110d$$

Let's gather the terms involving \(a\) on one side and the terms involving \(d\) on the other side. Subtract \(18a\) from both sides:

$$72d = 22a - 18a + 110d$$

$$72d = 4a + 110d$$

Now, subtract \(110d\) from both sides:

$$72d - 110d = 4a$$

$$-38d = 4a$$

We can divide both sides by 2:

$$-19d = 2a$$

This gives us a relationship between the first term (\(a\)) and the common difference (\(d\)): \(2a = -19d\).

Now, we need to find the sum of the first 20 terms (\(S_{20}\)). Using the formula for the sum of the first \(n\) terms with \(n=20\):

$$S_{20} = \frac{20}{2}[2a + (20-1)d]$$

$$S_{20} = 10[2a + 19d]$$

We already found that \(2a = -19d\). We can substitute this into the expression for \(S_{20}\):

$$S_{20} = 10[-19d + 19d]$$

$$S_{20} = 10[0]$$

$$S_{20} = 0$$

Thus, the sum of the first 20 terms of the AP is 0.

Step-by-Step Calculation Summary

This result shows that for this specific Arithmetic Progression, the sum of the first 20 terms is zero.

Revision Table: Key Concepts of AP Sum

Additional Information on AP Sum Properties

In an Arithmetic Progression, the sum of terms equidistant from the beginning and the end is constant. This property is embedded in the sum formula \(S_n = \frac{n}{2}(a_1 + a_n)\), where the sum is the average of the first and last term multiplied by the number of terms.

In this problem, the condition \(S_9 = S_{11}\) is crucial. It implies a specific relationship between the first term and the common difference. The fact that \(S_{20} = 0\) is a direct consequence of this relationship. It tells us that the terms in the AP balance out, with positive and negative terms cancelling each other out perfectly up to the 20th term.

Consider the terms from the 10th to the 11th. Since \(S_{11} = S_9\), this means the sum of the 10th and 11th terms must be zero:

\(S_{11} - S_9 = a_{10} + a_{11}\)

If \(S_{11} = S_9\), then \(a_{10} + a_{11} = 0\).

The 10th term is \(a + (10-1)d = a + 9d\).

The 11th term is \(a + (11-1)d = a + 10d\).

So, \((a + 9d) + (a + 10d) = 0\), which simplifies to \(2a + 19d = 0\).

Now, consider the sum of the first 20 terms:

\(S_{20} = \frac{20}{2}(2a + (20-1)d) = 10(2a + 19d)\)

Since we found that \(2a + 19d = 0\), substituting this into the \(S_{20}\) formula gives:

\(S_{20} = 10(0) = 0\)

This confirms our previous calculation and provides an alternative way to see why the sum of the first 20 terms is 0, directly from the sum of the 10th and 11th terms being 0.