Understanding Arithmetic Progression (AP) Terms

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by \(d\). The first term of an AP is usually denoted by \(a\).

The formula for the \(n\)th term of an AP is given by:

\(a_n = a + (n-1)d\)

where:

\(a_n\) is the \(n\)th term

\(a\) is the first term

\(n\) is the term number

\(d\) is the common difference

Finding the (p + q)th Term of the AP

Using the formula for the \(n\)th term, we can find the \((p+q)\)th term by replacing \(n\) with \((p+q)\):

\(a_{p+q} = a + ((p+q)-1)d\)

\(a_{p+q} = a + (p+q-1)d\)

Finding the (p – q)th Term of the AP

Similarly, we can find the \((p-q)\)th term by replacing \(n\) with \((p-q)\) in the formula for the \(n\)th term. We assume \(p > q\) for the term number to be positive, or that the term number can be non-positive if the sequence extends in both directions, but for standard AP problems, \(p-q\) is usually a positive integer index.

\(a_{p-q} = a + ((p-q)-1)d\)

\(a_{p-q} = a + (p-q-1)d\)

Calculating the Sum of the (p + q)th and (p – q)th Terms

Now, we need to find the sum of the \((p+q)\)th term and the \((p-q)\)th term. Let's add the expressions we found:

Sum = \(a_{p+q} + a_{p-q}\)

Sum = \([a + (p+q-1)d] + [a + (p-q-1)d]\)

Let's remove the brackets and group like terms:

Sum = \(a + (p+q-1)d + a + (p-q-1)d\)

Sum = \((a + a) + (p+q-1)d + (p-q-1)d\)

Sum = \(2a + d[(p+q-1) + (p-q-1)]\)

Now, let's simplify the expression inside the square brackets:

\((p+q-1) + (p-q-1) = p + q - 1 + p - q - 1\)

The \(+q\) and \(-q\) terms cancel each other out:

\(p - 1 + p - 1 = 2p - 2\)

Substitute this back into the sum expression:

Sum = \(2a + d(2p-2)\)

Sum = \(2a + 2d(p-1)\)

We can factor out a 2 from the expression:

Sum = \(2[a + d(p-1)]\)

Now, let's look at the expression inside the square brackets: \(a + (p-1)d\). This is exactly the formula for the \(p\)th term of the AP, \(a_p\).

So, the sum is:

Sum = \(2 \times a_p\)

This means the sum of the \((p+q)\)th term and the \((p-q)\)th term of an AP is equal to twice the \(p\)th term.

Verifying the AP Sum Result

The derivation shows algebraically that \(a_{p+q} + a_{p-q} = 2 a_p\). This is a general property of arithmetic progressions, stemming from the fact that the \(p\)th term is the arithmetic mean of terms equidistant from it.

For example, in an AP with \(a=1, d=2\), the terms are 1, 3, 5, 7, 9, 11, 13, ...

Let \(p=4\) and \(q=2\). Then \(p+q=6\) and \(p-q=2\).

The \(p\)th term (\(4\)th term) is \(a_4 = 7\). Twice the \(p\)th term is \(2 \times 7 = 14\).

The \((p+q)\)th term (\(6\)th term) is \(a_6 = 11\).

The \((p-q)\)th term (\(2\)nd term) is \(a_2 = 3\).

The sum of the \((p+q)\)th and \((p-q)\)th terms is \(a_6 + a_2 = 11 + 3 = 14\).

The sum \(14\) is indeed equal to twice the \(p\)th term \(14\). This example supports the derived result.

Conclusion on Sum of AP Terms

The sum of the \((p+q)\)th and \((p-q)\)th terms of an AP simplifies to \(2a + 2(p-1)d\), which is exactly twice the formula for the \(p\)th term, \(a + (p-1)d\). Therefore, the sum is equal to twice the \(p\)th term.

Revision Table: Key AP Concepts

Additional Information: Properties of AP

Arithmetic Progressions have several interesting properties:

If three terms are in AP, say \(x, y, z\), then the middle term is the arithmetic mean of the other two: \(y = \frac{x+z}{2}\), or \(2y = x+z\). Our result \(a_{p+q} + a_{p-q} = 2a_p\) is a generalization of this property. The term \(a_p\) is exactly in the middle of \(a_{p-q}\) and \(a_{p+q}\) in terms of their position number (\(p-q\), \(p\), \(p+q\)). The position \(p\) is the average of positions \(p-q\) and \(p+q\): \(\frac{(p-q)+(p+q)}{2} = \frac{2p}{2} = p\).

Adding or subtracting a constant to each term of an AP results in another AP with the same common difference.

Multiplying or dividing each term of an AP by a non-zero constant results in another AP. The new common difference is the original common difference multiplied or divided by the constant.

The sum of terms equidistant from the beginning and end of a finite AP is constant and equal to the sum of the first and last terms.