Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.4 | Set 1

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

Solution:

(i) 50, 46, 42, …. to 10 terms

From the given A.P. we get

n(total number of terms) = 10

a(first term) = a₁ = 50

d(Common Difference) = a₂ – a₁ = 46 – 50 = -4

Now put all these values in the given formula

S = n/2 (2a + (n – 1) d)

We, get

S = 10/2 (100 + (9) (-4))

= 5 (100 – 36)

= 5 (64)

= 320

Hence, the sum of the given A.P. = 320.

(ii) 1, 3, 5, 7, … to 12 terms

From the given A.P. we get

n = 12

a =  a₁ = 1

d = a₂ – a₁ = 3 – 1 = 2

Now put all these values in the given formula

S = n/2 (2a + (n – 1) d)

we get

S = 12/2 (2(1) + (12-1) (2))

= 6 (2 + (11) (2))

= 6 (2 + 22)

= 6 (24)

= 144

Hence, the sum of the given A.P. = 144.

(iii) 3, 9/2, 6, 15/2, … to 25 terms

From the given A.P. we get

n = 25

a = a₁ = 3

d = a₂ – a₁ = 9/2 – 3 = 3/2

Now put all these values in the given formula

S = n/2 (2a + (n – 1) 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

Hence, the sum of the given A.P. = 525.

(iv) 41, 36, 31, … to 12 terms

From the given A.P. we get

n = 12

a = a₁ = 41

d = a₂ – a₁ = 36 – 41 = -5

Now put all these values in the given formula

S = n/2 (2a + (n – 1) d)

We get

S = 12/2 (2(41) + (12 – 1) (-5))

= 6 (82 + (11) (-5))

= 6 (82 – 55)

= 6 (27)

= 162

Hence, the sum of the given AP = 162.

(v) a+b, a-b, a-3b, … to 22 terms

From the given A.P. we get

n = 22

a = a₁ = a+b

d = a₂ – a₁ = (a – b) – (a + b) = -2b

Now put all these values in the given formula

S = n/2 (2a + (n – 1) 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

Hence, the sum of the given AP = 22a – 440b.

(vi) (x – y)², (x²+ y²), (x + y)², … to n terms

From the given A.P. we get

n = n

a = a₁ = (x – y)²

d = a_{2 –} a₁ = (x² + y²) – (x – y)² = 2xy

Now put all these values in the given formula

S = n/2 (2a + (n – 1) 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)

Hence, the sum of the given AP = n (x² + y² – 3xy + xyn).

(vii) (x – y)/(x + y), (3x – 2y)/(x + y), (5x – 3y)/(x + y), … to n terms

From the given A.P. we get

n = n

a = a₁ = (x – y)/(x + y)

d = a_{2 –} a₁ = (3x – 2y)/(x + y) – (x-y)/(x+y) = (2x – y)/(x+y)

Now put all these values in the given formula

S = n/2 (2a + (n – 1) 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}

Hence, the sum of the given A.P. = 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 + b) + (a + b)² + s…. + [(a + b)² + 6ab]

Solution:

(i) 2 + 5 + 8 + … + 182

From the given A.P. we get

a(first term) = a₁ = 2

d(common difference) = a2 – a1 = 5 – 2 = 3

a_n = 182

Find the value of n using the given formula

a_n = a + (n – 1) d

182 = 2 + (n – 1) 3

182 = 2 + 3n – 3

182 = 3n -1

3n = 182 + 1

n = 183/3

= 61

Now, we find the sum of the given A.P. using the following formula

S = n/2 (a + l)

= 61/2 (2 + 182)

= 61/2 (184)

= 61 (92)

= 5612

Hence, the sum of the series = 5612

(ii) 101 + 99 + 97 + … + 47

From the given A.P. we get

a = a₁ = 101

d = a₂ – a₁ = 99 – 101 = -2

a_n = 47

Find the value of n using the given formula

a_n = a + (n-1) d

47 = 101 + (n-1)(-2)

47 = 101 – 2n + 2

2n = 103 – 47

2n = 56

n = 56/2 = 28

Now, we find the sum of the given A.P. using the following formula

S = n/2 (a + l)

= 28/2 (101 + 47)

= 28/2 (148)

= 14 (148)

= 2072

Hence, the sum of the series = 2072

(iii) (a – b)² + (a² + b²) + (a + b)² + s…. + [(a + b)² + 6ab]

From the given A.P. we get

a = a₁ = (a-b)²

d = a_{2 –} a₁ = (a² + b²) – (a – b)² = 2ab

a_n = [(a + b)² + 6ab]

Find the value of n using the given formula

a_n = 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 – a^{2 –} b² + 2ab + 2ab = 2abn

12ab = 2abn

n = 12ab / 2ab

= 6

Now, we find the sum of the given A.P. using the following formula

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)

Hence, the sum of the series = 6 (a² + b² + 3ab)

Question 3. Find the sum of first n natural numbers.

Solution:

Let AP be 1, 2, 3, 4, …, n

So, from the given A.P. we get

a(first term) = a₁ = 1

d(common difference) = a₂ – a₁ = 2 -1 = 1

l = n

Now, we find the sum of the given A.P. using the following formula

S = n/2 [2a + (n – 1) d]

= n/2 [2(1) + (n – 1) 1]

= n/2 [2 + n – 1]

= n/2 [n – 1]

Hence, the sum of the first n natural numbers is n(n – 1)/2

Question 4. Find the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5

Solution:

According to the question the natural numbers that are divisible by 2 or 5 are

2 + 4 + 5 + 6 + 8 + 10 + … + 100 = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95)

So, the A.P = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95) 

Now lets take (2 + 4 + 6 +…+ 100)

From this sequence we get

a = 2, d = 4 – 2 = 2, a_n = 100

Find the value of n using the given formula

a_n = a + (n – 1)d

100 = 2 + (n – 1)2

100 = 2 + 2n – 2

2n = 100

n = 100/2

= 50

Now, we find the sum of the given A.P. using the following formula

S = n/2 (2a + (n – 1)d)

= 50/2 (2(2) + (50 – 1)2)

= 25 (4 + 49(2))

= 25 (4 + 98)

= 2550

Now we take (5 + 15 + 25 +…+95)

From this sequence we get

a = 5, d = 15 – 5 = 10, a_n = 95

Find the value of n using the given formula

a_n = a + (n – 1)d

95 = 5 + (n – 1)10

95 = 5 + 10n – 10

10n = 95 +10 – 5

10n = 100

n = 100/10

= 10

Now, we find the sum of the given A.P. using the following formula

S = n/2 (2a + (n – 1)d)

= 10/2 (2(5) + (10 – 1)10)

= 5 (10 + 9(10))

= 5 (10 + 90)

= 500

Hence, the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5 = 2550 + 500 = 3050

Question 5. Find the sum of first n odd natural numbers.

Solution:

According to the question 

A.P = 1, 3, 5, 7……n

So, from the given A.P. we get

a = 1, d = 3 – 1 = 2, n = n

Now, we find the sum of the given A.P. using the following formula

S = n/2 [2a + (n – 1)d]

= n/2 [2(1) + (n – 1)2]

= n/2 [2 + 2n – 2]

= n/2 [2n]

= n²

Hence, the sum of the first n odd natural numbers = n².

Question 6. Find the sum of all odd numbers between 100 and 200

Solution:

According to the question 

A.P = 101, 103, 105, …, 199

So, from the given A.P. we get

a = 101, d = 103 – 101 = 2, a_n = 199

Find the value of n using the given formula

a_n = a + (n – 1)d

199 = 101 + (n – 1)2

199 = 101 + 2n – 2

2n = 199 – 101 + 2

2n = 100

n = 100/2

= 50

Now, we find the sum of the given A.P. using the following formula

S = n/2[a + l]

= 50/2 [101 + 199]

= 25 [300]

= 7500

Hence, the sum of the all odd numbers between 100 and 200 = 7500.

Question 7. Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

Solution:

According to the question

A.P = 3, 9, 15,…,999

So, from the given A.P. we get

a = 3, d = 9 – 3 = 6, a_n = 999

Find the value of n using the given formula

a_n = a + (n – 1)d

999 = 3 + (n – 1)6

999 = 3 + 6n – 6

6n = 999 + 6 – 3

6n = 1002

n = 1002/6

= 167

Now, we find the sum of the given A.P. using the following formula

Sum of n terms, S = n/2 [a + l]

= 167/2 [3 + 999]

= 167/2 [1002]

= 167 [501]

= 83667

Hence, the sum of all odd integers between 1 and 1000 which are divisible by 3 = 83667.

Question 8. Find the sum of all integers between 84 and 719, which are multiples of 5

Solution:

According to the question

A.p. = 85, 90, 95, …, 715

So, from the given A.P. we get

a = 85, d = 90 – 85 = 5, a_n = 715

Find the value of n using the given formula

a_n = a + (n – 1)d

715 = 85 + (n – 1)5

715 = 85 + 5n – 5

5n = 715 – 85 + 5

5n = 635

n = 635/5

= 127

Now, we find the sum of the given A.P. using the following formula

Sum of n terms, S = n/2 [a + l]

= 127/2 [85 + 715]

= 127/2 [800]

= 127 [400]

= 50800

Hence, the sum of all integers between 84 and 719, which are multiples of 5 = 50800.

Question 9. Find the sum of all integers between 50 and 500 which are divisible by 7

Solution:

According to question

A.P = 56, 63, 70, …, 497

So, from the given A.P. we get

a = 56, d = 63 – 56 = 7, an = 497

Find the value of n using the given formula

a_n = a + (n – 1)d

497 = 56 + (n – 1)7

497 = 56 + 7n – 7

7n = 497 – 56 + 7

7n = 448

n = 448/7

= 64

Now, we find the sum of the given A.P. using the following formula

Sum of n terms, S = n/2 [a + l]

= 64/2 [56 + 497]

= 32 [553]

= 17696

Hence, the sum of all integers between 50 and 500 which are divisible by 7 = 17696.

Question 10. Find the sum of all even integers between 101 and 999.

Solution:

According to question

AP = 102, 104, 106, …, 998

So, from the given A.P. we get

a = 102, d = 104 – 102 = 2, a_n = 998

Find the value of n using the given formula

a_n = a + (n – 1)d

998 = 102 + (n – 1)(2)

998 = 102 + 2n – 2

2n = 998 – 102 + 2

2n = 898

n = 898/2

= 449

Now, we find the sum of the given A.P. using the following formula

Sum of n terms, S = n/2 [a + l]

= 449/2 [102 + 998]

= 449/2 [1100]

= 449 [550]

= 246950

Hence, the sum of all even integers between 101 and 999 = 246950.