The document provides examples of arithmetic progressions and their patterns. It defines an arithmetic progression as a series of terms where the difference between consecutive terms is constant. It gives examples of arithmetic progressions with varying first terms and common differences. It also provides formulas for calculating the sum of the first n terms of an arithmetic progression based on the first term, last term, and common difference.

3. We now look
for some
patterns
which occur
in our daily
life.
Such
examples
are:-
(i) Reena applied
for a job with starting
monthly salary of Rs
8000,with an annual
increment of 500 in
her salary, her salary
for the 1st, 2nd,3rd….will
be respectively
8500,9000,9500……

11. Let us denote the first term of an AP by
a1, second term by a2,……,nth term by
an and the common difference by d.
Then the AP becomes a1,a2,a3…..,an.
So, a2 –a1=a3- a2=….=an-an-1=d.

15. •Similarly, when
a=-7,d=-2,
the AP is -7,-9,-11, -13…
a=1.0,d=0.1, the AP
is 1.0, 1.1,1.2,1.3,…
a=0,d=1x1/2, the
AP is 0, 1x1/2,3,4x1/2,6,….
a= 2,d=0,the AP is
2,2,2,2,…….

19. The sum of the first n terms of an AP
is given by S = 2/n[2n + (n – 1) d]
We can also write this as S = 2/n[a
+ a + (n-1)d] i.e., S = n/2 (a + an)
Now , if there are only n terms in an
AP, then an = 1, the last term. From
(3), we see that S = n/ (a + 1)