# 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, â€¦. to 10 terms**

**(ii) 1, 3, 5, 7, â€¦ to 12 terms**

**(iii) 3, 9/2, 6, 15/2, â€¦ to 25 terms**

**(iv) 41, 36, 31, â€¦ to 12 terms**

**(v) a+b, a-b, a-3b, â€¦ to 22 terms**

**(vi) (x â€“ y) ^{2}, (x^{2}+ y^{2}), (x + y)^{2}, â€¦ to n terms**

**(vii) (x â€“ y)/(x + y), (3x â€“ 2y)/(x + y), (5x â€“ 3y)/(x + y), â€¦ to n terms**

**Solution:**

(i)50, 46, 42, â€¦. to 10 termsFrom the given A.P. we get

n(total number of terms) = 10

a(first term) = a

_{1}= 50d(Common Difference) = a

_{2}– a_{1}= 46 – 50 = -4Now 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, â€¦ to 12 termsFrom the given A.P. we get

n = 12

a = a

_{1}= 1d = a

_{2}– a_{1}= 3 – 1 = 2Now 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, â€¦ to 25 termsFrom the given A.P. we get

n = 25

a = a

_{1}= 3d = a

_{2}– a_{1}= 9/2 – 3 = 3/2Now 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, â€¦ to 12 termsFrom the given A.P. we get

n = 12

a = a

_{1}= 41d = a

_{2}– a_{1 }= 36 – 41 = -5Now 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, â€¦ to 22 termsFrom the given A.P. we get

n = 22

a = a

_{1}= a+bd = a

_{2}– a_{1}= (a – b) – (a + b) = -2bNow 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 â€“ y)^{2}, (x^{2}+ y^{2}), (x + y)^{2}, â€¦ to n termsFrom the given A.P. we get

n = n

a = a

_{1}= (x – y)^{2}d = a

_{2 – }a_{1}= (x^{2}+ y^{2}) – (x – y)^{2}= 2xyNow 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 (x

^{2}+ y^{2}– 2xy) + 2xyn – 2xy)= n/2 Ã— 2 ((x

^{2}+ y^{2}– 2xy) + xyn – xy)= n (x

^{2}+ y^{2 }– 3xy + xyn)Hence, the sum of the given AP = n (x

^{2}+ y^{2}– 3xy + xyn).

(vii)(x – y)/(x + y), (3x – 2y)/(x + y), (5x – 3y)/(x + y), â€¦ to n termsFrom the given A.P. we get

n = n

a = a

_{1}= (x – y)/(x + y)d = a

_{2 – }a_{1}= (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 + â€¦ + 182**

**(ii) 101 + 99 + 97 + â€¦ + 47**

**(iii) (a – b) ^{2} + (a^{2} + b) + (a + b)^{2} + sâ€¦. + [(a + b)^{2} + 6ab]**

**Solution:**

(i)2 + 5 + 8 + â€¦ + 182From the given A.P. we get

a(first term) = a

_{1}= 2d(common difference) = a2 – a1 = 5 – 2 = 3

a

_{n}= 182Find the value of n using the given formula

a

_{n}= a + (n – 1) d182 = 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 + â€¦ + 47From the given A.P. we get

a = a

_{1}= 101d = a

_{2}– a_{1}= 99 – 101 = -2a

_{n}= 47Find the value of n using the given formula

a

_{n}= a + (n-1) d47 = 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}+ (a^{2}+ b^{2}) + (a + b)^{2}+ sâ€¦. + [(a + b)^{2}+ 6ab]From the given A.P. we get

a = a

_{1}= (a-b)^{2}d = a

_{2 –}a_{1}= (a^{2}+ b^{2}) – (a – b)^{2}= 2aba

_{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)2aba

^{2}+ b^{2}+ 2ab + 6ab = a^{2}+ b^{2}â€“ 2ab + 2abn – 2aba

^{2}+ b^{2}+ 8ab – a^{2 –}b^{2}+ 2ab + 2ab = 2abn12ab = 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 (a

^{2}+ b^{2}– 2ab + a^{2}+ b^{2}+ 2ab + 6ab)= 3 (2a

^{2}+ 2b^{2}+ 6ab)= 3 Ã— 2 (a

^{2}+ b^{2}+ 3ab)= 6 (a

^{2}+ b^{2}+ 3ab)Hence, the sum of the series = 6 (a

^{2}+ b^{2}+ 3ab)

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

**Solution:**

Let AP be 1, 2, 3, 4, â€¦, n

So, from the given A.P. we get

a(first term) = a

_{1}= 1d(common difference) = a

_{2}– a_{1}= 2 -1 = 1l = 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 + â€¦ + 100 = (2 + 4 + 6 +â€¦+ 100) + (5 + 15 + 25 +â€¦+95)

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

Now lets take (2 + 4 + 6 +â€¦+ 100)

From this sequence we get

a = 2, d = 4 – 2 = 2, a

_{n}= 100Find the value of n using the given formula

a

_{n}= a + (n – 1)d100 = 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 +â€¦+95)

From this sequence we get

a = 5, d = 15 – 5 = 10, a

_{n }= 95Find the value of n using the given formula

a

_{n}= a + (n – 1)d95 = 5 + (n – 1)10

95 = 5 + 10n â€“ 10

10n = 95 +10 â€“ 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, 7â€¦â€¦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]

= n

^{2}Hence, the sum of the first n odd natural numbers = n

^{2}.

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

**Solution:**

According to the question

A.P = 101, 103, 105, â€¦, 199

So, from the given A.P. we get

a = 101, d = 103 – 101 = 2, a

_{n}= 199Find the value of n using the given formula

a

_{n}= a + (n – 1)d199 = 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,â€¦,999

So, from the given A.P. we get

a = 3, d = 9 – 3 = 6, a

_{n}= 999Find the value of n using the given formula

a

_{n}= a + (n – 1)d999 = 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, â€¦, 715

So, from the given A.P. we get

a = 85, d = 90 – 85 = 5, a

_{n}= 715Find the value of n using the given formula

a

_{n}= a + (n – 1)d715 = 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, â€¦, 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)d497 = 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, â€¦, 998

So, from the given A.P. we get

a = 102, d = 104 – 102 = 2, a

_{n}= 998Find the value of n using the given formula

a

_{n}= a + (n – 1)d998 = 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.