PDF homework

In the previous theory, we understood that Arithmetic Progress $$A.P$$ formed a sequence of $$a$$, $$a + d$$, $$a + 2d$$, $$a + 3d$$, $$a + 4d$$, $$a + 5d$$. Here each number is called a term.

The first term is '$$a$$', the second term is '$$a + d$$' which is obtained by adding the difference $$(d)$$, and the third term is '$$a+2d$$'.

The terms of an $$A.P.$$ can be written several ways. Now let's see the few ways which are:
General $$n^t$$$$^h$$ term:
When  $$n ∈ N$$, $$n = 1, 2, 3, 4, ...$$

$$t_1 = a = a + (1 - 1) d$$

$$t_2 = a + d = a + (2 - 1) d$$

$$t_3 = a + 2d = a + (3 - 1) d$$

$$t_4 = a + 3d = a + (4 - 1) d$$

Here '$$t$$' refers to terms, and '$$n$$' denotes the number of terms.

In general, the $$n^t$$$$^h$$ term denoted by $$t_n$$ can be written as $$t_n = a + (n - 1) d$$.
In a finite $$A.P.$$ whose first term is '$$a$$' and last term '$$l$$', then the number of terms in the $$A.P.$$ is given by $l=a+\left(n-1\right)d⇒n=\left(\frac{l-a}{d}\right)+1$
Common difference:
To find the common difference of an $$A.P$$., we should generally subtract the first term from the second term, the second from the third and so on.
The first term $$t_1 = a$$ and the second term $$t_2 = a + d$$.

Difference between $$t_1$$ and $$t_2$$ is $$t_2 - t_1 = (a + d) - a = d$$.

Similarly, $$t_2 = a + d$$ and $$t_3 = a + 2d$$.

Therefore, $$t_3 - t_2 = a + 2d - a + d = d$$.

So, in general $$d = t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = t_5 - t_4$$.

Thus$d={t}_{n}-{t}_{n-1}$ where $$n = 1, 2, 3, …$$
The common difference of an $$A.P.$$ can be positive, negative, or zero.
Example:
1. Consider an $$A.P.$$ $$10, 13, 16, 19, 22, ...$$

$$d = t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = t_5 - t_4$$.

$$d = 13-10 = 16-13 = 19-16 = 22-19 =3$$.

Here the common difference is $$3$$.

2. Take an $$A.P.$$ $$-7, -10, -13, -16, ...$$

$$d = t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = t_5 - t_4$$.

$$d = -10-(-7) = -13-(-10) = -16-(-13) = -3$$.

Here the common difference is $$-3$$.

3. If an $$A.P$$ is $$-7, -7, -7, -7, -7, ...$$

$$d = t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = t_5 - t_4$$.

$$d = -7-(-7) = -7-(-7) = -7-(-7) = -7-(-7) =0$$.

Here the common difference is $$0$$.

An arithmetic progression having a common difference of zero is called a constant arithmetic progression. For example, here the $$A.P$$ $$-7, -7, -7, -7, -7, ...$$ is called constant arithmetic progression.
Condition for three numbers to be in $$A$$.$$P$$.
If $$a$$, $$b$$, $$c$$ are in $$A$$.$$P$$. then $$a = a$$, $$b = a +d$$, $$c = a +2d$$

So, $$a + c$$ $$= 2a + 2d = 2 (a + d) = 2b$$

Thus, $$2b = a + c$$

Similarly, if $$2b = a +c$$, then $$b − a = c −b$$ so $$a, b, c$$ are in $$A.P.$$

Thus three non-zero numbers $$a, b, c$$ are in $$A$$.$$P$$. if and only if $$2b = a + c$$
Example:
If 3$$+$$ $$x$$18$$-$$ $$x$$, 5$$x$$ $$+$$ 1 are in $$A$$.$$P$$. then find $$x$$.

Since the given $$A$$.$$P$$. series has three numbers, we can use the above-derived expression $$2b = a + c$$.

Let us take $\begin{array}{l}\underset{⏟}{3+x},\underset{⏟}{18-x},\underset{⏟}{5x+1}\\ \phantom{\rule{1.176em}{0ex}}a\phantom{\rule{2.058em}{0ex}}b\phantom{\rule{2.646em}{0ex}}c\phantom{\rule{0.882em}{0ex}}\end{array}$

Now substitute the known values in the expression $$2b = a + c$$.

$\begin{array}{l}2\left(18-x\right)=3+x+5x+1\\ \\ 36-2x=4+6x\\ \\ 36-4=6x+2x\\ \\ 32=8x\\ \\ \frac{32}{8}=x\\ \\ x=4\end{array}$

The value of $$x$$ $$=$$ 4

Important!
Key takeaways:
• The common difference of an $$A.P.$$ can be positive, negative, or zero.
• The common difference of constant $$A.P.$$ is zero.
• If '$$a$$' and '$$l$$' are the first and last terms of an $$A.P.$$ then the number of terms $$(n)$$ is $n=\left(\frac{l-1}{d}\right)+1$