arithmetic progressions
Post on 07-Jul-2015
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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)
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