Fundamental concepts of A.P. — lesson. Mathematics CBSE, Class 10.

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In the previous theory, we obtain the basic ideas of Arithmetic progression. Let's discuss some more concept for a better understanding with examples.

First, we see how the arithmetic operations \(( +, -, ×, ÷)\) behaves on \(A.P.\)

1. Addition & Subtraction:

In an Arithmetic Progression, if every term is added or subtracted by a constant, then the resulting sequence is also an \(A.P.\)

Example:

1. Consider the \(A.P.\) \(20, 25, 30, 35, 40, 45....\)

Here, \(a = 20, d = 25 - 20 = 5\)

In the above sequence, each term is added constantly by \(5\), which makes the sequence as \(A.P.\)

2. Take the same \(A.P.\) \(20, 25, 30, 35, 40, 45....\)

Subtract by the common difference \(5\) constantly throughout the sequence. And the resultant sequence also makes an \(A.P.\)

That is, \(15, 20, 25, 30, 35, 40....\)

2. Multiplication & Division:

If every term is multiplied or divided by a non-zero number in an Arithmetic Progression, then the resulting sequence is also an\(A.P.\)

Example:

Consider the \(A.P\) \(20, 25, 30, 35, 40, 45....\)

Here, \(a = 20\) and \(d = 5\)

If we multiply each term in the \(A.P\) by a constants non-zero number, say \(5\), we get the sequence which is also an \(A.P\).

That is \((20 × 5), (25 × 5), (30 × 5), (35 × 5), (40 × 5), (45 × 5) ....\)

The \(A.P\) is \(100, 125, 150, 175, 190, 215....\).

Similarly, if we divide it by \(5\), the resulting sequence also forms an \(A.P.\).

Let's see how it is done.

\begin{array}{l} = \frac{20}{5}, \frac{25}{5}, \frac{30}{5}, \frac{35}{5}, \frac{40}{5}, \frac{45}{5} \\ = 4, 5, 6, 7, 8, 9 \end{array}

The sequence \(4, 5, 6, 7, 8, 9....\) is also an \(A.P\).

3. Sum of the consecutive terms:

If the sum of three consecutive terms of an \(A.P.\) is given, then they can be taken as \(a - d\), \(a\) and \(a + d\) and the common difference is \(d\).

Example:

If the sum of three consecutive terms of an \(A.P.\) is \(24\). And the common difference is \(4\), obtain the \(A.P.\) sequence.

Solution:

The sum of three consecutive terms of an \(A.P.\) is \(24\).

Common difference \( d = 4\).

Then the \(A.P\) is \(a - d\), \(a\), \(a + d\).

So, \(a - d + a + a + d = 24\)

\( 3a = 24\)

\(a = 8\)

That is, \(8 - 4\), \(8\), \(8 + 4\).

Hence, the \(A.P\) is \(4, 8, 12\)...

If the sum of four consecutive terms of an \(A.P.\) is given then, they can be taken as \(a - 3d\), \(a - d\), \(a + d\) and \(a + 3d\). Here the common difference is \(2d\).

Example:

Find \(A.P.\), when the sum of four consecutive terms of an \(A.P.\) is \(96\), and the common difference is \(6\).

Solution:

The sum of four consecutive terms of an \(A.P.\) is \(96\).

Common difference \(2d = 12\)

\(d = 6\).

So, the \(A.P.\) is \(a - 3d\), \(a - d\), \(a + d\) and \(a + 3d\)

The sum of the \(A.P.\) is \(a - 3d + a - d + a + d + a + 3d = 96\)

\(4a = 96\)

\(a = 24\)

Now we know the values of \(a\) and \(d\).

So, \(24 - 3(6)\), \(24 - 6\), \(24 + 6\) and \(24 + 3(6)\)