The document defines and provides examples of arithmetic progressions (AP). It states that an AP is a sequence where the difference between consecutive terms is constant. The key characteristics of an AP include: - The first term is denoted as a - The common difference is denoted as d - The nth term is calculated as an = a + (n-1)d Several examples are provided to illustrate how to determine if a sequence is an AP and how to calculate individual terms and the sum of terms in an AP using the above formulae.

1. Arithmetic
Progression
Class 10
Chapter 5
Prepared by:
Sharda Chauhan
TGT Mathematics

2. Sequence: A list of numbers
having specific relation
between the consecutive
termsisgenerally called a
sequence.
e.g. 1, 3, 5, 7,… … … (next term
to a term isobtained by
adding 2 with it)
& 2, 6, 18, 54,… … .( next term
to a term isobtained by
multiplying 3 with it)

3. Arithm
etic P
rogression: If varioustermsof a
sequence are formed by adding a fixed
number to the previous term or the
difference between two successive
termsisa fixed number, then the sequence
iscalled AP.
e.g.1) 2, 4, 6, 8, … … … the sequence of even
numbersisan example of AP
2) 5, 10, 15, 20, 25…..
In thiseach term isobtained by adding 5 to
the preceding term except first term.

4. I
llustrativeexample forA.P.
=
d,where d=1
a a+
d a+
2d a+
3d……… … … …

5. The general form of an Arithmetic Progression
is
a , a +
d , a +2d , a +3d … … … … … …, a +(n-
1)d
Where ‘a’ is first term and
‘d’ iscalled common difference.

6. Common Difference - The fixed number which
is obtained by subtracting any term of AP from
its previous term.
If we take
First term of an AP asa
and Common Difference
asd,
Then,
nth term of that AP will be
An =a +(n-1)d

7. Notice in thissequence that if we find the difference
between any term and the term before it we always
get 4.
4 isthen called the common difference and is
denoted with the letter d.
To get to the next term in the sequence
we would add 4 so a recursive formula
for this sequence is:
an  an1  4
The first term in the sequence would be a1 whic h
issometimes just written as a.
3, 7, 11, 15, 19 … d =4 a =3

8. +4 +4 +4 +4
3, 7, 11, 15, 19 …
d =4 a =3
Each time you want another term in the sequence you’d add d. Thiswould
mean the sec ond term wasthe first termplusd.T
he third term isthe first term
plusd plusd (added twic e). T
he fourth termisthe first termplusd plusd plusd
(added three times). S
o you c an see to get the nth term we’d take the first
term and add d (n -1)times.
an  a  n 1d
T
rythisto get
the 5th term.
a5  3 514  316 19

9. Let’ssee an example!!
Let a=
2, d=
2, n=
12,find
AnAn=a+(n-1)d
=2+(12-1)2
=2+(11)2
=2+22
Therefore, An
=24
Hence solved.

10. Tocheck thata given termisin A.P. or not.
2, 6, 10, 14….
Here first term a =2,
find differences in the next terms
a2-a1 =6 –2 =4
a3-a2 =10 –6 =4
a4-a3 =14 –10 =4
Since the differencesare
common.
Hence the given termsare in A.P.

11. Problem : Find the value ofk forwhich
the given series isin A.P. 4, k –1, 12
Solution : Given A.P. is4, k –1 , 12… ..
If series is A.P. then the differences will be
common.
d1 =d1
a2 –a1 = a3 –a2
k –1 –4 =12 –(k –1)
k –5 = 12 –k +1
k +k =12 +1 +5

12. The sumofn term
s, we find as,
Sum =n X [(first term +last term) / 2]
Now last term will be =a +(n-1) d
Therefore,
Sum(Sn
) =n X [{
a +a +(n-1) d }/2 ]
=n/2 [2a +(n+1)d]

13. DERIVATION
T
he sum to n termsisgiven by:
Sn =a +(a +d) +(a +2d) +… +(a +(n –1)d) (1)
Ifwe write this out backwards, we get:
Sn =(a +(n –1)d) +(a +(n –2)d) +… +a (2)
Now let’sadd (1) and (2):
2Sn =[2a +(n –1)d] +[2a +(n –1)d] +…
… … … +[2a +(n –1)d]
So, Sn = n/2[2
a +(n –1)d]

14. Problem . Find number of terms of
A.P.100, 105, 110, 115,………500
Solution:
First term is a = 100 , an =500
Common difference is d = 105 -100 =5
nth term is an =a +(n-1)d
500 =100 +(n-1)5
500 - 100 =5(n –1)
400 =5(n –1)
5(n –1) =400

15. 5(n –1) =400
n –1 =400/5
n - 1 = 80
n =80 +1
n =81
Hence the no. of termsare 81.

16. Problem . Find the sum of 30 terms of given
A.P. ,12 , 20 , 28 , 36………
Solution :Given A.P. is12 , 20, 28 , 36
Its first term is a =12
Common difference isd =20 –12 =8
The sum to n termsof an arithmetic progression
Sn =n/2 [ 2a +(n - 1)d ]
=½x 30 [ 2x 12 +(30-1)x 8]
=15 [ 24 +29 x8]

17. =15[24 +232]
=15 x 246
=3690
THE SUM OF TERMS IS 3690

18. Problem . Find the sum of terms in given A.P.
2 , 4 , 6 , 8 , ……………… 200
Solution:Its first term is a =2
Common difference isd =4 –2 =2
nth term is an =a +(n-1)d
200 =2 +(n-1)2
200 - 2 =2(n –1)
2(n –1) =198
n –1 =99, n =100

19. The sum to n terms of an arithmetic
progression Sn =n/2[ 2a +(n - 1)d ]
S100 =100/2 [ 2x 2 +(100-1)x 2]
=50 [ 4 +198]
=50[202]
=10100

20. T
he difference betw
een tw
o termsofan
APcan be formulated asbelow:-
nthterm –kthterm
=t(n)–t(k)
={
a +(n-1)d}–{a +(k-1)d }
=a +nd –d –a –kd +d =nd –kd
Hence,
t(n)–t(k) =(n –k)d

21. Any Questions?
ASK!
Thank You!