3. DEFINATIONS
Arithmetic progression(AP) = An AP is a list of numbers in which each
term is obtained by adding a fixed numbers to preceding term except the
first term. Remember that it can be positive, negative, or zero.
E.G = Reena applied for a job and got selected. She has been offered a
job with a starting monthly salary of RS8000, with annual increment
of RS500 in her salary. Her salary for the 1st,2nd,3rd,….years will be
respectively
8000,8500,9000,9500,…………..
Common Difference = The fixed numbers is called common difference, it
is denoted as “d”.
a2 - a1 = a3 - a2 = …. = an- an-1 = d
4. Sequence = A sequence is an arrangement of numbers in a
definite order following definite rule.
Finite Sequence = A sequence containing definite numbers of
terms.
Infinite Sequence = A sequence is called infinite sequence if it
is not a finite sequence.
General Form Of An AP = a,a+d,a+2d,a+3d represents an AP
where a is the first term and d is the common difference. This
is called general form of an AP.
5. DERIVATION OF FORMULA FOR nth TERM OF AN AP
Let us consider the general form of an AP
a,a+d,a+2d,a+3d…….nth term
1st term = a1
a+(1-1)d = a+0d
2nd term = a2
a+(2-1)d = a+1d
a+d
3rd term = a3
a+(3-1)d = a+ 2d
a+2d
4th term = a4
a+(4-1)d = a+3d
a+3d
--------An term ( last
term)
a
------------a+(n-1)d =
--------------a+(n-1)d
This formula can be used it find nth term of an AP whose first term is a
and common difference is d
an=a+(n-1)d
6. Let us consider some examples
1) Find the 10th term of an
AP:2,7,12,…?
Sol:- here a=2, d=a2-a1 = 72=5, n=10
an=a+(n-1)d
a10=2+(10-1)5
=2+(9)5
= 2+45
a10 = 47
2)How many two digit no. are
divisible by 3?
Sol:- AP: 12,15,18……99
Here a=12, d=15-12=3,
an= 99
an=a+(n-1)d
99=12+(n-1)3
87=(n-1)3
n-1 = 87/3 = 29
n=29+1
n=29
7. SUM OF FIRST n TERMS OF AN AP
Let a be the first term and d be the common difference of an
AP .
Let sn denote the sum of first n terms and l be its last term.
Sn = 1+2+3+4+5
Sn = 5+4+3+2+1
2Sn = 6+6+6+6+6
2Sn = 30
Sn = 30/2 = 15
Sn = 15
8. DERIVATION OF FORMULA
a,a+d,a+2d……..
The nth term of this AP is a a+(n-1)d. let s denote the sum
of first n terms of the AP.
S=a+(a+d)+(a+2d)+…+[a+(n-1)d]
Rewriting the terms in reversing order,
2s=[2a+(n-1)d]+[2a+(n-1)d]+…+[2a+(n-1)d]+[2a+(n-1)d]
n times
2S=n[2a+(n-1)d]
S=n/2[2a+(n-1)d]
9. The Sum Of The First n Terms Of An AP Is Given By
S=n/2[2a+(n-1)d]
S=n/2[a+a+(n-1)d]
S=n/2(a+an)
If there is only n terms in AP then an=l, the last
term from (3)
S=n/2(a+l)
