Arithmetic progressions

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……

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.

•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,…….

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)