Let a and d be real numbers. Then the numbers of the form a, a + d , a + 2d , a + 3d , a + 4d , ... is said to form Arithmetic Progression denoted by A.P. The number ‘a’ is called the first term and ‘d’ is called the common difference.

**Arithmetic
Progression**

Let us begin with the following two illustrations

Make the following
figures using match sticks

(i) How many match
sticks are required for each figure? 3,5,7 and 9.

(ii) Can we find the
difference between the successive numbers?

5 − 3 = 7 − 5 = 9 − 7 =
2

Therefore, the
difference between successive numbers is always 2.

A man got a job whose
initial monthly salary is fixed at ₹10,000 with an annual increment of ₹2000.
His salary during 1*st* , 2*nd* and 3*rd* years will be ₹ 10000,
₹ 12000 and ₹ 14000 respectively.

If we now calculate the
difference of the salaries for the successive years, we get 12000 – 10000 =
2000; 14000 – 12000 = 2000 . Thus the difference between the successive numbers
(salaries) is always 2000.

Did you observe the
common property behind these two illustrations? In these two examples, the
difference between successive terms always remains constant. Moreover, each
term is obtained by adding a fixed number (2 and 2000 in illustrations 1 and 2
presented above) to the preceding term except the first term. This fixed number
which is a constant for the differences between successive terms is called the “common difference”.

**Definition**

Let *a* and *d*
be real numbers. Then the numbers of the form *a*, *a* + *d* , *a*
+ 2*d* , *a *+* *3*d *,* a *+* *4*d *, ...
is said to form* *Arithmetic Progression* *denoted by A.P. The* *number ‘*a*’ is
called the first
term and ‘*d*’ is
called the common
difference.

Simply, an Arithmetic
Progression is a sequence whose successive terms differ by a constant number.
Thus, for example, the set of even positive integers 2, 4, 6, 8, 10, 12,… is an
A.P. whose first term is *a* = 2 and common difference is also *d* =
2 since 4 − 2 = 2,
6 − 4 = 2, 8 − 6 = 2, …

Most of common real−life
situations often produce numbers in A.P.

Note

·
The difference between any two consecutive terms of an A.P. is
always constant. That** **constant value is called the common difference.

·
If there are finite numbers of terms in an A.P. then it is
called Finite Arithmetic** **Progression. If there are infinitely many terms
in an A.P. then it is called Infinite Arithmetic Progression.

1. The terms of an A.P. can
be written as

*t*_{1}* *=* a *=* a *+*
*(1* *−1)*d *,* t*_{2}* *=* a *+* d *=*
a *+* *(2* *−1)*d *,* t*_{3}* *=* a* +* *2*d
*=* a *+* *(3* *−1)*d *,* t*_{4}* *=*
a *+* *3*d *=* a *+* *(4* *−1)*d *, . . .

In general, the *n*^{th}
term denoted by *t _{n}* can be written as

2. In general to find
the common difference of an A.P. we should subtract first term from the second
term, second from the third and so on.

For example, *t*_{1}
= *a*, *t* _{2} = *a* +*d*

Therefore, *t* _{2}
− *t*_{1} = (*a* + *d* ) −*a* = *d*

Similarly, *t*
_{2} = *a* +*d*, *t *_{3 }=* a *+* *2*d*,...

Therefore, *t* _{3}
− *t* _{2} = (*a* + 2*d*) −(*a* + *d* )
= *d*

In general, *d* = *t*
_{2} −*t*_{1 }= *t* _{3} −*t* _{2}
= *t* _{4} −*t*_{3} = ....

So, *d* = *t _{n}* −

Let us try to find the
common differences of the following A.P.’s

(i) 1, 4, 7, 10,….

*d*= 4 − 1 = 7 − 4 = 10 − 7
= ... = 3

(ii) 6, 2, − 2, −6,…

*d *=* *2* *−* *6*
*= −* *2* *−* *2* *= −6* *−* *(−2)* *=*
*...* *= −4

**Thinking Corner:** If *t _{n}* is the

Check whether the
following sequences are in A.P. or not?

(i) *x* + 2, 2*x* + 3, 3*x* +
4,….

(ii)
2, 4, 8, 16,...

(iii) 3√2, 5√2, 7√2, 9√2,...

To check that the given
sequence is in A.P., it is enough to check if the differences** **between the consecutive
terms are equal or not.

(i)
*t *_{2}* *-*t*_{1}*
*=* *(2*x *+* *3)* *−* *(*x *+* *2)* *=*
x *+* *1* *

*t *_{3}* *-*t*_{2}*
*= (3*x *+*
*4)* *−* *(2*x *+* *3)* *=* x *+* *1

*t *_{2}* *- *t*_{1}*
*=* t *_{3}* *− *t*_{2}

Thus, the differences
between consecutive terms are equal.

Hence the sequence *x*
+ 2, 2*x* + 3, 3*x* + 4,... is in A.P.

(ii)
*t *_{2}* *-*t*_{1}*
*=* *4* *−* *2* *=* *2

*t *_{3}* *-*t*_{2}*
*=* *8* *−* *4* *=* *4

*t *_{2}* *-*t*_{1}*
*=* t *_{3}* *−*t*_{2}

Thus, the differences between consecutive terms are not equal. Hence the terms of the sequence 2, 4, 8, 16, . . . are not in A.P.

(iii)
*t _{2} -t_{1} = 5√2 −
3√ 2 = 2√ 2*

*t _{3} -t_{2} = 7√2 − 5√ 2 = 2√ 2*

*t _{4} -t_{3} = 9√ 2 − 7√2 = 2√ 2*

Thus, the differences
between consecutive terms are equal. Hence the terms of the sequence 3*√*2,
5*√*2, 7*√*2, 9*√*2,... are in A.P

Write an A.P. whose
first term is** **20** **and common difference is** **8.

First term** **=

Arithmetic Progression
is *a*, *a* + *d*, *a* + 2*d* , *a* + 3*d*,...

In this case, we get 20,
20 + 8, 20 + 2(8), 20 + 3(8),...

So, the required A.P. is
20, 28, 36, 44,…..

**Note **

An Arithmetic
progression having a common difference of zero is called a constant** **arithmetic
progression.

Find the** **15* ^{th}*,

We have, first term** **=

We know that *n ^{th
}* term (general term) of an A.P.
with first term

*t _{n} *=

*t*_{15}* *=* a *+* *(15*
*−1)*d *=* a *+* *14*d *=* *3* *+* *14*
*(12)* *=* *171

(Here *a* = 3 and *d*
= 12)

*t*_{24}= *a* + (24 −1)*d* = *a* + 23*d* = 3 +23(12) =
279

The *n ^{th}* (general
term) term is given by

*t _{n} *=

Thus,* t _{n} *= 3 + (

*t _{n} *= 12

**Note**

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* gives n = ( [*l-a*] / d)
+1.

Find the number of terms
in the A.P.**
**3,** **6,** **9,** **12,…,** **111.

First term *a *=*
*3* *; common difference *d* =
6 − 3 = 3 ; last term l = 111

Thus the A.P. contain 37
terms.

Determine the general
term of an A.P. whose** **7^{th}** **term is −1** **and** **16^{th}** **term** **is 17.

Let the A.P. be* **t*_{1}** **,

It is given that *t*_{7} = −1 and *t*_{16} = 17

*a + (7 −1)d = −1 *and *a +
(16 −1)d = 17*

*a + 6d *= −1* ...(1)*

*a + 15d* = 17* **...(2)*

Subtracting equation (1) from equation (2), we get 9*d* = 18
gives *d* = 2

Putting *d* = 2 in
equation (1), we get *a* + 12 =
−1 so *a* = –13

Hence, general term *t _{n}*
=

= −13 + (*n* −1)×2
= 2*n* −15

If** ***l ^{th}*

(i) *x* (*m* −
*n* ) + *y* (*n* − *l* ) + *z* (*l* − *m*) =
0 (ii) (*x* − *y* )*n* + (*y* − *z* )*l* + (z− *x*
)*m* = 0

(i) Let* **a*** **be the first term and

*t*_{l} = x, tm = y, *t*_{n} = *z*

Using the general term
formula

*a + (l −1)d = x **...(1)*

*a + (m −1)d = y** ...(2)*

*a + (n −1)d = z **...(3)*

We have, *x* (*m*
− *n* ) + *y*(*n* − *l* ) + *z* (*l* −*m*)

*=a *[(*m *−*n *)*
*+* *(*n *−* l *)* *+* *(*l *−*m *)]* *+*
d *[(*m *−*n*)(*l *−* *1)* *+* *(*n *−*
l *)(*m *−1)* *+* *(*l *−*m*)(*n *−1)]

*=a *[0]* *+*d*[*lm
*−* ln *−*m *+* n *+* mn *−* lm *−*n *+* l *+*ln
*−* mn *−*l *+* m*]

*=a*(0)* *+* *d(0)* *=* *0

(ii) On subtracting
equation (2) from equation (1), equation (3) from equation (2) and equation (1) from equation (3), we get

*x *−* y *=* *(*l
*−*m*)*d*

y− *z* = (*m*
−*n*)*d*

*z *−*x *=* *(*n
*−*l*)*d*

(*x* − *y* )*n*
+ (*y* − *z* )*l* + (*z* − *x*)*m* = [(*l* −*m*)*n*
+ (*m* −*n* )*l* + (*n* −*l*)*m* ]*d* = *ln*
− *mn* + *lm* − *nl* + *nm* − *lm d* = 0

**Note**

In an Arithmetic
Progression

·
If every term is added or subtracted by a constant, then the
resulting sequence is also** **an A.P.

·
If every term is multiplied or divided by a non-zero number,
then the resulting sequence** **is also an A.P.

·
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 *.
Here the common difference is* d*.

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

In an A.P., sum of four
consecutive terms is** **28** **and their sum of their** **squares is 276. Find the four numbers.

Let us take the four
terms in the form** **(

Since sum of the four
terms is 28,

*a *−* *3*d *+*a
*−* d *+*a *+* d *+*a *+* *3*d *=* *28

4*a* = 28 gives *a*
= 7

Similarly, since sum of
their squares is 276,

(*a* − 3d)^{2} + (a − d )^{2} + (a + d )^{2} + (a + 3d)^{2} = 276.

a^{2} − 6ad + 9d^{2}
+ a^{2} − 2ad + d^{2} +a^{2} + 2ad +d^{2} + a^{2}
+ 6ad + 9d^{2} = 276

4*a*^{2} + 20d^{2} =276 ⇒
4(7)^{2} + 20d^{2}
= 276.

*d*^{2} = 4 gives *d* = ± 2

If *d* = 2 then the
four numbers are 7 - 3(2), 7 – 2, 7 + 2, 7+3(2)

That is the four numbers
are 1, 5, 9 and 13.

If *a* = 7, *d*
= −2 then the four numbers are 13, 9, 5 and 1

Therefore, the four
consecutive terms of the A.P. are 1, 5, 9 and 13.

**Condition for three
numbers to be in A.P.**

If a, b, c are in A.P. then b = a +d, c = a + 2d

so a + c = 2a + 2d

2(a + d) = 2b

Thus 2b

**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**

A mother devides ₹207** **into three parts such
that the amount are in A.P.** **and gives it to her three children. The product of the two least
amounts that the children had ₹4623. Find the amount received by each child.

Let the amount received
by the three children be in the form of A.P. is given by* *

*a *–*d *,* a*,*
a *+* d *. Since, sum of the amount is ₹207, we have

(*a* − *d* ) +*a*
+ (*a* +*d*) = 207

3*a* = 207 gives *a*
= 69

It is given that product
of the two least amounts is 4623.

(*a* − *d* )*a*
= 4623

(69 − *d* )69 =
4623

*d *=* *2

Therefore, amount given
by the mother to her three children are

₹(69−2), ₹69, ₹(69+2).
That is, ₹67, ₹69 and ₹71.

Tags : Definition, Theorem, Illustration, Terms, Common Difference, Example, Solution | Mathematics , 10th Mathematics : UNIT 2 : Numbers and Sequences

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10th Mathematics : UNIT 2 : Numbers and Sequences : Arithmetic Progression | Definition, Theorem, Illustration, Terms, Common Difference, Example, Solution | Mathematics

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