Arithmetic Progression (A.P) | Mathematics (Maths) Class 11 - Commerce PDF Download

Sequence and Series

A. Sequence

A  sequence  is  a set of terms in a  definite  order with a rule for obtaining the terms.

e.g. 1 , 1/2 , 1/3 , ....... , 1/n , ........ is a sequence .

A sequence is a function whose domain is the set N of natural numbers. Since the domain for every sequence is the set of N natural numbers, therefore a sequence is represented by its range. f : N → R,                      then f(n) = t_n n ε N is called a sequence and is denoted by

{f(1), f(2), f(3),................} =

{t₁, t₂, t₃,.........................} = {t_n}

Real Sequence : A sequence whose range is a subset of R is called a real sequence.

Examples :

(i) 2, 5, 8, 11, ...........

(ii) 4, 1, -2, -5, ............

(iii) 3, -9, 27, -81, ...........

Types of Sequence : On the basis of the number of terms there are two types of sequence.

(i) Finite sequence : A sequence is said to be finite if it has finite number of terms.

(ii) Infinite sequence : A sequence is said to be infinite if it has infinite number of terms.

Series : By adding or substracting the terms of a sequence, we get an expression which is called a series. If a₁, a₂, a₃, ......., a_n is a sequence, then the expression a₁ + a₂ + a₃+ ...... + a_n is a series.

Example.

(i) 1 + 2 + 3 + 4 + ............ + n

(ii) 2 + 4 + 8 + 16 + ..............

Progression : It is not necessary that the terms of a sequence always follow a certain pattern or they are described by some explicit formula of the n^th term. Those sequences whose terms follow certain patterns are called progressions.

Ex.1 Write down the sequence whose nth term is 

(i) (-1)ⁿ

(ii)

Sol. (i) Let a_n = (-1)ⁿ. Putting = 1, 2, 3, 4,...... successively, we get

a₁ = (-1)¹ = - 1  

a₂ = (-1)² = 8/5

a₃ = (-1)³ = -11/5   

a₄ = (-1)⁴ = 14/5

..........................................

Hence we obtain the sequence -1, 8/5, -11/5, 14/5,......

(ii) Let a_n = . Putting n = 1, 2, 3, 4,... successively, we get

a₁ =

a₂ =

a₃ =  = 0 

a₄ =

...........................................

Hence we obtain the sequence /2, /8, 0, -/32,...

Ex.2 If sum of n terms of a sequence is given by S_n = 2n² + 3n, find its 50^th term.

Sol. Let t_n is n^th term of the sequence so t_n = s_n - s_{n - 1} = 2n² + 3n - 2(n - 1)² - 3 (n - 1) = 4n + 1

so t₅₀ = 201.

B. Arithmetic Progression (AP)

AP is a sequence whose terms increase or decrease by a fixed number . This fixed number is called the common difference . If a is the first term & d the common difference, then AP can be written as a , a + d , a + 2 d , ....... a + (n - 1) d , ....... .

n^th term of this AP

t_n = a + (n - 1) d, where d = a_n - a_n-1 .

The sum of the first n terms of the AP is given by ; S_n =  [2 a + (n - 1)d] = [a + l] .

where l is the last term.

Remarks : 

(i) If each term of an A.P. is increased, decreased, multiplied or divided by the same non zero number, then the resulting sequence is also an AP .

(ii) Three numbers in AP can be taken as a - d , a , a + d ; four numbers in AP can be taken as a - 3d, a - d, a + d, a + 3d ; five numbers in AP are a - 2d , a - d ,   a, a + d, a + 2d & six terms in AP are a - 5d, a - 3d, a - d, a + d, a + 3d, a + 5d etc.

(iii) The common difference can be zero, positive or negative .

(iv) The sum of the terms of an AP equidistant from the beginning  &  end  is constant and equal to the sum of first & last terms .

(v) Any term of an AP (except the first) is equal to half the sum of terms which are equidistant from it .

a_n = 1/2 (a_n-k + a_n+k), k < n .

For k = 1, a_n = (1/2) (a_n-1+ a_n+1) ; For k = 2, a_n = (1/2) (a_n-2+ a_n+2) and so on .

(vi) t_r = S_r - S_r-1

(vii) If a , b , c are in AP ⇒ 2 b = a + c.

Ex. 3 Find the sum of all the three digit natural numbers which on division by 7 leaves remainder 3.

Sol. All these numbers are 101, 108, 115,....... 997, to find n.

997 = 101 + (n - 1) 7   ⇒ n = 129 

so S = [101 + 997] = 70821.

Ex. 4 The sum of first three terms of an A.P. is 27 and the sum of their squares is 293 . Find the sum to 'n' terms of the A.P.

Sol. Let a - d , a , a + d be the numbers  ⇒ a = 9

Also (a - d)² + a² + (a + d)² = 293. ⇒ 3a² + 2d² = 293 ⇒ d² = 25 ⇒ d = ± 5

therefore numbers are 4, 9, 14.

Hence the A.P. is 4, 9, 14, ...... 

⇒ s_n = [5 n + 3] or 14, 9, 4, .....

⇒ s_n = [33 - 5 n]

Ex.5 Let a_n be the n^th term of an arithmetic progression. Let S_n be the sum of the first n terms of the arithmetic progression with a₁ = 1 and a₃ = 3a₈. Find the largest possible value of S_n.

Sol. From a₃ = 3a₈ we obtain 1 + 2d = 3(1 + 7d) ⇒ d = - 2/19.

Then S_n = = [19 - (n - 1)] = .

now consider 20n - n² = - [n² - 20n] = - [(n - 10)² - 100]

 S_n =  now, S_n will be maximum if n = 10 and (S_n)_max =

Ex.6 Suppose a₁, a₂,.... are in A.P. and S_k denotes the sum of the first k terms of this A.P. If S_n/S_m = n⁴/m⁴ for all m, n, ε N, then prove that

Sol. Putting a₁ = a, we have

Replacing n by 2n + 1 and m by 2m + 1, we get

⇒  

Ex.7 If the sum of m terms of an A.P. is equal to the sum of either the next n terms or the next p terms prove that (m + n)  = (m +p)

Sol. Let the A.P. be a, a + d, a + 2d,....

Given T₁ + T₂ + ... + T_m = T_{m + 1} + T_{m + 2} +....+T_{m + n}             ...(1)

Adding T₁ + T₂ + .... + T_m on both sides in (i), Then

2(T₁ + T₂+....+T_m) = T₁ + T₂ + ....+ T_m + T_{m + 1} +... +....+T_{m + n}           ⇒             2 S_m = S_{m + n}

∴ 2.  {2a + (m - 1)d} =  {2a + (m + n - 1) d}

Let 2a + (m - 1)d = x    

⇒  mx = {x + nd} ...(2)

⇒ (m - n) x = (m + n) nd 

again  T₁ + T₂ + .... + T_m = T_{m + 1} + T_{m + 2} +....+T_{m + p} ...(3)

Similarly (m - p) x = (m + p) pd

dividing (2) by (3), we get

⇒(m - n) (m + p) p = (m - p) (m + n) n

dividing both sides by mnp, we have (m + p)  = (m + n)

Hence (m + n)  = (m + p)

Ex.8 Show that any positive integral power (except the first) of a positive integer m, is the sum of m consecutive odd positive integers. Find the first odd integer for m^r (r > 1).

Sol. Let us find k such that m^r = (2k + 1) + (2k + 3) + .... + (2k + 2m - 1)

∴ m^r =  [2k + 1 + 2k + 2m - 1] 

⇒m^{r - 1} = 2k + m

Note that m^{r - 1} - m is an even integer for all r, m ε N and r > 1. Therefore, k = (m^{r - 1} - m)/2 is an integer. Thus, the first term is given by m^{r - 1}-m + 1.