Terms and common difference of an A.P. — lesson. Mathematics CBSE, Class 10.

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In the previous theory, we understand that Arithmetic Progress

$A.P$ form a sequence of

$a, a + d, a + 2 d, a + 3 d, a + 4 d, a + 5 d$ . 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

$l = a + (n - 1) d \Rightarrow n = (\frac{l - a}{d}) + 1$

Common difference:

To find the common difference of an

$A.P$ generally, we should 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

$A$ .

$P$ .

$a$ ,

$b$ ,

$c$ 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

$\begin{array}{l} \underset{⏟}{3 + x}, \underset{⏟}{18 - x}, \underset{⏟}{5 x + 1} \\ a b c \end{array}$

Now substitute the known values in the expression

$2b = a + c$ .

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

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 $A.P.$ is zero.
If ' $a$ ' and ' $l$ ' are the first and last terms of an $A.P.$ then the number of terms $(n)$ is $n = (\frac{l - 1}{d}) + 1$

$1$

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2. Terms and common difference of an A.P.

Theory: