Arithmetic Progression (A.P)

Definition: A sequence (progression) is a set of numbers in a definite order with a definite rule of obtaining the numbers.

Arithmetic Progression (A. P.)

Definition: An A.P. is a sequence whose terms increase or decrease by a fixed number , called the common difference of the A.P.

nth Term

If a is the first term and d the common difference, the A.P. can be written as a , a + d , a + 2d , ……

The nth term an is given by Tn = a + (n − 1)d

Sum of n Terms

The sum Sn of the first n terms of such an A.P. is given by

$ \displaystyle S_n = \frac{n}{2}[2a + (n-1)d ] $

$ \displaystyle S_n = \frac{n}{2}[a + l ] $

where l is the last term (i.e. the nth term of the A.P.)

Notes:

* If a fixed number is added (subtracted) to each term of a given A.P. then the resulting sequence is also an A.P. with the same common difference as that of the given A.P.

* If each term of an A.P. is multiplied by a fixed number (say k) (or divided by a non-zero fixed number), the resulting sequence is also an A.P. with the common difference multiplied by k.

* If a1, a2, a3…..and b1, b2, b3…are two A.P.’s with common differences d and d’ respectively then a1+b1, a2+b2, a3+b3 , …is also an A.P. with common difference d + d’

* If we have to take three terms in an A.P., it is convenient to take them as a − d, a, a + d.
In general , we take a − rd , a − (r − 1)d ,……a − d , a , a + d ,…….a + rd in case we have to take (2r + 1) terms in an A.P

* If we have to take four terms , we take a − 3d , a − d , a + d , a + 3d.
In general, we take a − (2r − 1)d, a − (2r − 3)d,….a − d , a + d,…..a + (2r − 1)d, in case we have to take 2r terms in an A.P.

* If a1, a2, a3, ….. an are in A.P. then a1 + an = a2 + an−1 = a3 + an−2 = . . . . . and so on.

* If nth term of any sequence is a linear expression in n, then the sequence is an AP, whose common difference is the coefficient of n.

* If sum of n terms of any sequence is a quadratic in n, whose constant term is zero, then the sequence is an AP, whose common difference is twice the coefficient of n2. If the constant term is non-zero, then it is an A.P. from second term onwards.

* If {tn} is an A.P., then the common difference, d, is given by

$ \displaystyle d = \frac{t_p – t_q}{p-q} $ , (p , q ∈ N ) $

Arithmetic Mean(s):

* If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two i.e. if a, b, c are in A.P. then b = (a + c)/2 is the A.M. of a and c.

* If a1, a2, … an are n numbers then the arithmetic mean (A) of these numbers is

$ \displaystyle A = \frac{1}{n} (a_1 + a_2 + a_3 + —- + a_n) $

* The n numbers A1, A2…..An are said to be A.M.’ s between the numbers a and b if a, A1, A2 ,…..An, b are in A.P. If d is the common difference of this A.P.

then b = a + (n + 2 − 1)d

(b-a) = (n+1)d

$ \displaystyle d = \frac{b-a}{n+1} $

$ \displaystyle A_r = a + r(\frac{b-a}{n+1}) $

where Ar is the rth mean

Illustration : Let {tn} is an A.P. If t1 = 20 , tp = q , tq = p , find the value of m such that sum of the first m terms of the A.P. is zero.

Solution: Given, t1 = 20 , tp = q , tq = p

Common difference = d

$ \displaystyle \frac{t_p -t_q }{p-q} = \frac{q-p}{p-q} = -1 $

Let Sm = 0

$ \displaystyle \frac{m}{2}[2\times 20 + (m-1)(-1)] = 0 $

40 − m + 1 = 0

m = 41

Illustration : Find the number of terms in the series 20 , 58/3 , 56/3 , …. of which the sum is 300 , explain the double answer.

Solution: Clearly here a = 20 , d = −2/3

Let Sn = 300

$ \displaystyle \frac{n}{2}[2\times 20 + (n-1)(-2/3)] = 300 $

Simplifying

n2 − 61n + 900 = 0

⇒ n = 25 or 36.

Since common ratio is negative and S25 = S36 = 300 , it shows that the sum of the last eleven terms i.e. T26 , T27 , ……,… T36 is zero.

Illustration : n arithmetic means are inserted in between x and 2y and then between 2x and y. In case the rth mean in each case be equal, then find the ratio x/y

Solution:

Let A.P be a , x, x, ………, x, b

b = Tn + 2

b = a + (n + 1)d

$ \displaystyle d = \frac{b-a}{n+1} $

xr = Tr + 1

= a + r d

$ \displaystyle = a + r (\frac{b-a}{n+1} ) $

Now, putting a = x and b = 2y

and then again put a = 2x and b = y

and equate the results as the two means are equal.

$ \displaystyle \frac{x(n-r+1)+2yr}{n+1} = \frac{2x(n-r+1)+ yr}{n+1} $

$ \displaystyle \frac{x}{y} = \frac{r}{n-r+1} $

Exercise 1:

(i) If the angles of a triangle are in A.P. and tangent of the smallest angle is 1 then find all the angles of the triangle.

(ii) If a1, a2, a3, a4 , a5 , a6 are in A.P., then prove that the system of equations

a1x + a2y = a, a4x + a5y = a6 is consistent.

(iii) Let Sn denote the sum upto n terms of an A.P. If Sn = n2P and Sm = m2P , where m, n and p are positive integers and m ≠ n, then find Sp .

(iv) If s1, s2 and s3 are the sum of first n, 2n, 3n terms respectively of an arithmetical progression , then show that s3 = 3(s2 − s1).

(v) Let a1, a2, a3 , …… be an A.P. Prove that

$ \displaystyle \sum_{n=1}^{2m}(-1)^{n-1}a_n^2 = \frac{m}{2m-1}(a_1^2 – a_{2m}^2) $

Also Read :

Geometric Progression(G.P)
Arithmetico Geometric Progression(G.P)
Harmonic Progression(H.P)
Miscellaneous Progression
INEQUALITIES
Solved Problems : Progression & Series

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