RD Chapter 19 Arithmetic Progressions Ex 19.4 Solutions
Question - 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
Answer - 1 : -
(i) 50, 46, 42, …. to 10terms
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 valuesof ‘a’ and ‘d’, we get
S = 10/2 (100 + (9)(-4))
= 5 (100 – 36)
= 5 (64)
= 320
∴ The sum of the givenAP is 320.
(ii) 1, 3, 5, 7, … to 12terms
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 valuesof ‘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 givenAP is 144.
(iii) 3, 9/2, 6, 15/2, … to25 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 valuesof ‘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 givenAP is 525.
(iv) 41, 36, 31, … to 12terms
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 valuesof ‘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 givenAP is 162.
(v) a+b, a-b, a-3b, … to22 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 valuesof ‘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 givenAP 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 valuesof ‘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 givenAP 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 valuesof ‘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 givenAP is n/2(x+y) {n (2x-y) – y}
Question - 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]
Answer - 2 : -
(i) 2 + 5 + 8 + … + 182
First term, a = a1 =2
Common difference, d =a2 – a1 = 5 – 2 = 3
an termof 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 seriesis 5612
(ii) 101 + 99 + 97 + … + 47
First term, a = a1 =101
Common difference, d =a2 – a1 = 99 – 101 = -2
an termof 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 seriesis 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 termof given AP is [(a + b)2 + 6ab]
an = a+ (n-1) d
[(a +b)2 + 6ab] = (a-b)2 + (n-1)2ab
a2 + 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 seriesis 6 (a2 + b2 + 3ab)
Question - 3 : - Find the sum of first n natural numbers.
Answer - 3 : -
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 firstn natural numbers is n(n-1)/2
Question - 4 : - Find the sum of all – natural numbers between 1 and 100, which aredivisible by 2 or 5
Answer - 4 : -
The natural numberswhich 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 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 thenumbers divisible by 2 or 5 is: 2550 + 500 = 3050
Question - 5 : - Find the sum of first n odd natural numbers.
Answer - 5 : -
Given an AP of first nodd 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 firstn odd natural numbers is n2.
Question - 6 : - Find the sum of all odd numbers between 100 and 200
Answer - 6 : -
The series is 101,103, 105, …, 199
Let the number ofterms 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 oddnumbers between 100 and 200 is 7500.
Question - 7 : - Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667
Answer - 7 : -
The odd numbersbetween 1 and 1000 divisible by 3 are 3, 9, 15,…,999
Let the number ofterms 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 oddintegers between 1 and 1000 which are divisible by 3 is 83667.
Hence proved.
Question - 8 : - Find the sum of all integers between 84 and 719, which are multiples of 5
Answer - 8 : -
The series is 85, 90,95, …, 715
Let there be ‘n’ termsin 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 integersbetween 84 and 719, which are multiples of 5 is 50800.
Question - 9 : - Find the sum of all integers between 50 and 500 which are divisible by 7
Answer - 9 : -
The series of integersdivisible by 7 between 50 and 500 are 56, 63, 70, …, 497
Let the number ofterms 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 allintegers between 50 and 500 which are divisible by 7 is 17696.
Question - 10 : - Find the sum of all even integers between 101 and 999
Answer - 10 : -
We know that all evenintegers 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 evenintegers between 101 and 999 is 246950.