 UPSKILL MATH PLUS

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In the previous theory, we obtained the basic ideas of arithmetic progression. Let's discuss some more concepts with examples for a better understanding.

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

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 constant 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.

$\begin{array}{l}=\frac{20}{5},\frac{25}{5},\frac{30}{5},\frac{35}{5},\frac{40}{5},\frac{45}{5}\\ \\ =4,\phantom{\rule{0.147em}{0ex}}5,\phantom{\rule{0.147em}{0ex}}6,\phantom{\rule{0.147em}{0ex}}7,\phantom{\rule{0.147em}{0ex}}8,\phantom{\rule{0.147em}{0ex}}9\end{array}$

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 $$24 + 3(6)$$

Hence, the $$A.P.$$ is $$6, 18, 30, 42, ...$$