Reena's annual salary increases by Rs. 500 each year, so her salary follows an arithmetic progression. The document defines an arithmetic progression as a series where each term after the first term is calculated by adding a constant value, called the common difference, to the previous term. It provides examples of arithmetic progressions with different initial terms and common differences. Finally, it presents formulas for calculating the sum of the first n terms of an arithmetic progression based on the initial term, common difference, and number of terms.

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)