Learning

is

simple

and

fun!

Complete thousands of tasks, learn solutions and improve your grades!

Register now! Browse subjects### Theory:

In the previous theory, we understand that Arithmetic Progress \(A.P\) form 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 $l=a+(n-1)d\Rightarrow n=\left(\frac{l-a}{d}\right)+1$

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 \(A.P.\) \(-7, -10, -13, -16, ...\)

\(d = t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = t_5 - t_4\).

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

**3**. If an \(A.P\) is \(-7, -7, -7, -7, -7, ...\)

\(d = t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = t_5 - t_4\).

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

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 $\begin{array}{l}\underset{\u23df}{3+x},\underset{\u23df}{18-x},\underset{\u23df}{5x+1}\\ \phantom{\rule{1.176em}{0ex}}a\phantom{\rule{2.058em}{0ex}}b\phantom{\rule{2.646em}{0ex}}c\phantom{\rule{0.882em}{0ex}}\end{array}$

Now substitute the known values in the expression \(2b = a + c\).

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

**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 $n=\left(\frac{l-1}{d}\right)+1$