Arithmetic Progression , nth term , Sum of n terms , Arithmetic Mean , Solved Examples

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.

n^th 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 n^th term a_n is given by T_n = a + (n − 1)d

Sum of n Terms

The sum S_n 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, ….. a_n are in A.P. then a₁ + a_n = a₂ + a_n−1 = a₃ + a_n−2 = . . . . . and so on.

* If n^th 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 {t_n} 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, … a_n 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…..A_n are said to be A.M.’ s between the numbers a and b if a, A₁, A₂ ,…..A_n, 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 A_r is the r^th mean

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

Solution: Given, t₁ = 20 , t_p = q , t_q = p

Common difference = d

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

Let S_m = 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 S_n = 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 , x₁ , x₂ , ………, x_n , b

b = T_n + 2

b = a + (n + 1)d

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

x_r = T_r + 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 :

(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 = a₃ , a₄x + a₅y = a₆ is consistent.

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

(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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