Arithmetic progressions
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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.
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![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)](https://image.slidesharecdn.com/arithmeticprogressions-141117041518-conversion-gate02/85/Arithmetic-progressions-19-320.jpg)
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Arithmetic progressions
- 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)