In this section, we shall discuss problems based on the sum to n terms of an A.P. This exercise can be used as a model of reference by the students to improve their conceptual knowledge and understand the different ways used to solve the problems. Here, the exercise-wise problems are solved and explained in simple language by using various methods and steps. Students mainly require lots of practice to analyse the types of problems that would appear in the exam. Through regular practice, students can improve their logical approach to solving problems. RD Sharma Class 11 Maths Solutions PDF links are provided below, where students can easily download and start practising offline.
RD Sharma Solutions for Class 11 Maths Exercise 19.4 Chapter 19 – Arithmetic Progressions
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Access answers to RD Sharma Solutions for Class 11 Maths Exercise 19.4 Chapter 19 – Arithmetic Progressions
1. Find the sum of the following arithmetic progressions:
(i) 50, 46, 42, …. to 10 terms
(ii) 1, 3, 5, 7, … to 12 terms
(iii) 3, 9/2, 6, 15/2, … to 25 terms
(iv) 41, 36, 31, … to 12 terms
(v) a+b, a-b, a-3b, … to 22 terms
(vi) (x – y)2, (x2 + y2), (x + y)2, … to n terms
(vii) (x – y)/(x + y), (3x – 2y)/(x + y), (5x – 3y)/(x + y), … to n terms
Solution:
(i) 50, 46, 42, …. to 10 terms
n = 10
First term, a = a1 = 50
Common difference, d = a2 – a1 = 46 – 50 = -4
By using the formula,
S = n/2 (2a + (n – 1) d)
Substitute the values of ‘a’ and ‘d’, we get
S = 10/2 (100 + (9) (-4))
= 5 (100 – 36)
= 5 (64)
= 320
∴ The sum of the given AP is 320.
(ii) 1, 3, 5, 7, … to 12 terms
n = 12
First term, a = a1 = 1
Common difference, d = a2 – a1 = 3 – 1 = 2
By using the formula,
S = n/2 (2a + (n – 1) d)
Substitute the values of ‘a’ and ‘d’, we get
S = 12/2 (2(1) + (12-1) (2))
= 6 (2 + (11) (2))
= 6 (2 + 22)
= 6 (24)
= 144
∴ The sum of the given AP is 144.
(iii) 3, 9/2, 6, 15/2, … to 25 terms
n = 25
First term, a = a1 = 3
Common difference, d = a2 – a1 = 9/2 – 3 = (9 – 6)/2 = 3/2
By using the formula,
S = n/2 (2a + (n – 1) d)
Substitute the values of ‘a’ and ‘d’, we get
S = 25/2 (2(3) + (25-1) (3/2))
= 25/2 (6 + (24) (3/2))
= 25/2 (6 + 36)
= 25/2 (42)
= 25 (21)
= 525
∴ The sum of the given AP is 525.
(iv) 41, 36, 31, … to 12 terms
n = 12
First term, a = a1 = 41
Common difference, d = a2 – a1 = 36 – 41 = -5
By using the formula,
S = n/2 (2a + (n – 1) d)
Substitute the values of ‘a’ and ‘d’, we get
S = 12/2 (2(41) + (12-1) (-5))
= 6 (82 + (11) (-5))
= 6 (82 – 55)
= 6 (27)
= 162
∴ The sum of the given AP is 162.
(v) a+b, a-b, a-3b, … to 22 terms
n = 22
First term, a = a1 = a+b
Common difference, d = a2 – a1 = (a-b) – (a+b) = a-b-a-b = -2b
By using the formula,
S = n/2 (2a + (n – 1) d)
Substitute the values of ‘a’ and ‘d’, we get
S = 22/2 (2(a+b) + (22-1) (-2b))
= 11 (2a + 2b + (21) (-2b))
= 11 (2a + 2b – 42b)
= 11 (2a – 40b)
= 22a – 440b
∴ The sum of the given AP is 22a – 440b.
(vi) (x – y)2, (x2 + y2), (x + y)2, … to n terms
n = n
First term, a = a1 = (x-y)2
Common difference, d = a2 – a1 = (x2 + y2) – (x-y)2 = 2xy
By using the formula,
S = n/2 (2a + (n – 1) d)
Substitute the values of ‘a’ and ‘d’, we get
S = n/2 (2(x-y)2 + (n-1) (2xy))
= n/2 (2 (x2 + y2 – 2xy) + 2xyn – 2xy)
= n/2 × 2 ((x2 + y2 – 2xy) + xyn – xy)
= n (x2 + y2 – 3xy + xyn)
∴ The sum of the given AP is n (x2 + y2 – 3xy + xyn).
(vii) (x – y)/(x + y), (3x – 2y)/(x + y), (5x – 3y)/(x + y), … to n terms
n = n
First term, a = a1 = (x-y)/(x+y)
Common difference, d = a2 – a1 = (3x – 2y)/(x + y) – (x-y)/(x+y) = (2x – y)/(x+y)
By using the formula,
S = n/2 (2a + (n – 1) d)
Substitute the values of ‘a’ and ‘d’, we get
S = n/2 (2((x-y)/(x+y)) + (n-1) ((2x – y)/(x+y)))
= n/2(x+y) {n (2x-y) – y}
∴ The sum of the given AP is n/2(x+y) {n (2x-y) – y}
2. Find the sum of the following series:
(i) 2 + 5 + 8 + … + 182
(ii) 101 + 99 + 97 + … + 47
(iii) (a – b)2 + (a2 + b2) + (a + b)2 + s…. + [(a + b)2 + 6ab]
Solution:
(i) 2 + 5 + 8 + … + 182
First term, a = a1 = 2
Common difference, d = a2 – a1 = 5 – 2 = 3
an term of given AP is 182
an = a + (n-1) d
182 = 2 + (n-1) 3
182 = 2 + 3n – 3
182 = 3n – 1
3n = 182 + 1
n = 183/3
= 61
Now,
By using the formula,
S = n/2 (a + l)
= 61/2 (2 + 182)
= 61/2 (184)
= 61 (92)
= 5612
∴ The sum of the series is 5612
(ii) 101 + 99 + 97 + … + 47
First term, a = a1 = 101
Common difference, d = a2 – a1 = 99 – 101 = -2
an term of given AP is 47
an = a + (n-1) d
47 = 101 + (n-1)(-2)
47 = 101 – 2n + 2
2n = 103 – 47
2n = 56
n = 56/2 = 28
Then,
S = n/2 (a + l)
= 28/2 (101 + 47)
= 28/2 (148)
= 14 (148)
= 2072
∴ The sum of the series is 2072
(iii) (a – b)2 + (a2 + b2) + (a + b)2 + s…. + [(a + b)2 + 6ab]
First term, a = a1 = (a-b)2
Common difference, d = a2 – a1 = (a2 + b2) – (a – b)2 = 2ab
an term of given AP is [(a + b)2 + 6ab]
an = a + (n-1) d
[(a + b)2 + 6ab] = (a-b)2 + (n-1)2aba2 + b2 + 2ab + 6ab = a2 + b2 – 2ab + 2abn – 2ab
a2 + b2 + 8ab – a2 – b2 + 2ab + 2ab = 2abn
12ab = 2abn
n = 12ab / 2ab
= 6
Then,
S = n/2 (a + l)
= 6/2 ((a-b)2 + [(a + b)2 + 6ab])
= 3 (a2 + b2 – 2ab + a2 + b2 + 2ab + 6ab)
= 3 (2a2 + 2b2 + 6ab)
= 3 × 2 (a2 + b2 + 3ab)
= 6 (a2 + b2 + 3ab)
∴ The sum of the series is 6 (a2 + b2 + 3ab)
3. Find the sum of first n natural numbers.
Solution:
Let AP be 1, 2, 3, 4, …, n
Here,
First term, a = a1 = 1
Common difference, d = a2 – a1 = 2 – 1 = 1
l = n
So, the sum of n terms = S = n/2 [2a + (n-1) d]
= n/2 [2(1) + (n-1) 1]
= n/2 [2 + n – 1]
= n/2 [n – 1]
∴ The sum of the first n natural numbers is n(n-1)/2
4. Find the sum of all – natural numbers between 1 and 100, which are divisible by 2 or 5
Solution:
The natural numbers which are divisible by 2 or 5 are:
2 + 4 + 5 + 6 + 8 + 10 + … + 100 = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95)
Now, (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95) are AP with a common difference of 2 and 10.
So, for the 1st sequence => (2 + 4 + 6 +…+ 100)
a = 2, d = 4-2 = 2, an = 100
By using the formula,
an = a + (n-1)d
100 = 2 + (n-1)2
100 = 2 + 2n – 2
2n = 100
n = 100/2
= 50
So now, S = n/2 (2a + (n-1)d)
= 50/2 (2(2) + (50-1)2)
= 25 (4 + 49(2))
= 25 (4 + 98)
= 2550
Again, for the 2nd sequence, (5 + 15 + 25 +…+95)
a = 5, d = 15-5 = 10, an = 95
By using the formula,
an = a + (n-1)d
95 = 5 + (n-1)10
95 = 5 + 10n – 10
10n = 95 +10 – 5
10n = 100
n = 100/10
= 10
So now, S = n/2 (2a + (n-1)d)
= 10/2 (2(5) + (10-1)10)
= 5 (10 + 9(10))
= 5 (10 + 90)
= 500
∴ The sum of the numbers divisible by 2 or 5 is: 2550 + 500 = 3050
5. Find the sum of first n odd natural numbers.
Solution:
Given an AP of first n odd natural numbers whose first term a is 1, and common difference d is 3
The sequence is 1, 3, 5, 7……n
a = 1, d = 3-1 = 2, n = n
By using the formula,
S = n/2 [2a + (n-1)d]
= n/2 [2(1) + (n-1)2]
= n/2 [2 + 2n – 2]
= n/2 [2n]
= n2
∴ The sum of the first n odd natural numbers is n2.
6. Find the sum of all odd numbers between 100 and 200
Solution:
The series is 101, 103, 105, …, 199
Let the number of terms be n
So, a = 101, d = 103 – 101 = 2, an = 199
an = a + (n-1)d
199 = 101 + (n-1)2
199 = 101 + 2n – 2
2n = 199 – 101 + 2
2n = 100
n = 100/2
= 50
By using the formula,
The sum of n terms = S = n/2[a + l]
= 50/2 [101 + 199]
= 25 [300]
= 7500
∴ The sum of the odd numbers between 100 and 200 is 7500.
7. Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667
Solution:
The odd numbers between 1 and 1000 divisible by 3 are 3, 9, 15,…,999
Let the number of terms be ‘n’, so the nth term is 999
a = 3, d = 9-3 = 6, an = 999
an = a + (n-1)d
999 = 3 + (n-1)6
999 = 3 + 6n – 6
6n = 999 + 6 – 3
6n = 1002
n = 1002/6
= 167
By using the formula,
Sum of n terms, S = n/2 [a + l]
= 167/2 [3 + 999]
= 167/2 [1002]
= 167 [501]
= 83667
∴ The sum of all odd integers between 1 and 1000, which are divisible by 3 is 83667.
Hence proved.
8. Find the sum of all integers between 84 and 719, which are multiples of 5
Solution:
The series is 85, 90, 95, …, 715
Let there be ‘n’ terms in the AP
So, a = 85, d = 90-85 = 5, an = 715
an = a + (n-1)d
715 = 85 + (n-1)5
715 = 85 + 5n – 5
5n = 715 – 85 + 5
5n = 635
n = 635/5
= 127
By using the formula,
Sum of n terms, S = n/2 [a + l]
= 127/2 [85 + 715]
= 127/2 [800]
= 127 [400]
= 50800
∴ The sum of all integers between 84 and 719, which are multiples of 5 is 50800.
9. Find the sum of all integers between 50 and 500 which are divisible by 7
Solution:
The series of integers divisible by 7 between 50 and 500 are 56, 63, 70, …, 497
Let the number of terms be ‘n’
So, a = 56, d = 63-56 = 7, an = 497
an = a + (n-1)d
497 = 56 + (n-1)7
497 = 56 + 7n – 7
7n = 497 – 56 + 7
7n = 448
n = 448/7
= 64
By using the formula,
Sum of n terms, S = n/2 [a + l]
= 64/2 [56 + 497]
= 32 [553]
= 17696
∴ The sum of all integers between 50 and 500, which are divisible by 7, is 17696.
10. Find the sum of all even integers between 101 and 999
Solution:
We know that all even integers will have a common difference of 2.
So, AP is 102, 104, 106, …, 998
We know, a = 102, d = 104 – 102 = 2, an = 998
By using the formula,
an = a + (n-1)d
998 = 102 + (n-1)2
998 = 102 + 2n – 2
2n = 998 – 102 + 2
2n = 898
n = 898/2
= 449
By using the formula,
Sum of n terms, S = n/2 [a + l]
= 449/2 [102 + 998]
= 449/2 [1100]
= 449 [550]
= 246950
∴ The sum of all even integers between 101 and 999 is 246950.
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