RBSE Class 10 Maths Chapter 5 – Arithmetic Progression Important questions and solutions are available here. All these questions are provided by the subject experts to help the students in scoring maximum marks. The RBSE Class 10 important questions and solutions given at BYJU’S have detailed explanations and are prepared according to the guidelines of the board.

Chapter 5 of RBSE Class 10 has three exercises which deals with finding the first term, common difference, nth term and sum of first n terms of an AP. Students will also be tested with word problems to check whether the given situation forms an AP or not. Students will be able to answer all types of questions on these concepts once they practice these important questions thoroughly.

### RBSE Maths Chapter 5: Exercise 5.1 Textbook Important Questions and Solutions

**Question 1: Find the first term a and common difference d for the following AP.**

**(i) 6, 9, 12, 15,…**

**(ii) -7, -9, -11, -13,…**

**Solution:**

(i) 6, 9, 12, 15,…

First term = a = 6

Common difference = d = Second term – First term

= 9 – 6

= 3

(ii) -7, -9, -11, -13,…

First term = a = -7

Common difference = d = Second term – First term

= -9 – (-7)

= -9 + 7

= -2

**Question 2: Find the first term a and common difference d for the following AP.**

**(i) 3/2, 1/2, -1/2, -3/2,….**

**(ii) -1, 1/4, 2/3,…**

**Solution:**

(i) 3/2, 1/2, -1/2, -3/2,….

First term = a = 3/2

Common difference = d = Second term – First term

= (½) – (3/2)

= (1 – 3)/2

= -2/2

= -1

(ii) -1, 1/4, 2/3,…

First term = a = -1

Common difference = d = Second term – First term

= (¼) – (-1)

= (¼) + 1

= (1 + 4)/4

= 5/4

Question 3: If first term a and common difference d of the AP is given, then find the first four terms of that progression.

a = 1/3, d = 4/3

Solution:

Given,

a = 1/3, d = 4/3

Let us consider the AP be a_{1}, a_{2}, a_{3}, a_{4}, a_{5},…

a_{1} = a = ⅓

a_{2} = a_{1} + d = ⅓ + 4/3 =(1 + 4)/3 = 5/3

a_{3} = a_{2} + d = 5/3 + 4/3 = (5 + 4)/3 = 9/3 = 3

a_{4} = a_{3} + d = 3 + 4/3 = (9 + 4)/3 = 13/3

Therefore, the first four terms are: ⅓, 5/3, 3, 13/3.

**Question 4: If first term a and common difference d of the AP is given, then find the first four terms of that progression.**

**a = 11, d = -4**

**Solution:**

Given,

a = 11, d = -4

Let us consider the AP be a_{1}, a_{2}, a_{3}, a_{4}, a_{5},…

a_{1} = a = 11

a_{2} = a_{1} + d = 11 + (-4) = 11 – 4 = 7

a_{3} = a_{2} + d = 7 + (-4) = 7 – 4 = 3

a_{4} = a_{3} + d = 3 + (-4) = 3 – 4 = -1

Therefore, the first four terms are: 11, 7, 3, -1

**Question 5: If first term a and common difference d of the AP is given, then find the first four terms of that progression.**

**a = 20, d = -3/4**

**Solution:**

Given,

a = 20, d = -3/4

Let us consider the AP be a_{1}, a_{2}, a_{3}, a_{4}, a_{5},…

a_{1} = a = 20

a_{2} = a_{1} + d = 20 + (-¾) = (80 – 3)/4 = 77/4

a_{3} = a_{2} + d = 77/4 + (-¾) = (77 – 3)/4 = 74/4

a_{4} = a_{3} + d =(74/4) + (-¾) = (74 – 3)/4 = 71/4

Therefore, the first four terms are: 20, 77/4, 74/4, 71/4.

### RBSE Maths Chapter 5: Exercise 5.2 Textbook Important Questions and Solutions

**Question 6: Find 10th term of AP: 2, 7, 12,…**

**Solution:**

Given AP: 2, 7, 12,…

First term = a = 2

Common difference = d = 7 – 2 = 5

nth term of AP:

a_{n} = a + (n – 1)d

a_{10} = 2 + (10 – 1)(5)

= 2 + 9(5)

= 2 + 45

= 47

Therefore, 10th term is 47.

**Question 7: Find 18th term of AP: √2, 3√2, 5√2,…**

**Solution:**

Given AP: √2, 3√2, 5√2,…

First term = a = √2

Common difference = d = 3√2 – √2 = 2√2

nth term of AP:

a_{n} = a + (n – 1)d

a_{18} = √2 + (18 – 1)(2√2)

= √2 + (17) (2√2)

= √2 + 34√2

= 35√2

Therefore, 18th term is 35√2.

**Question 8: Find 24th term of AP: 9, 13, 17, 21,…**

**Solution:**

Given AP: 9, 13, 17, 21,…

First term = a = 9

Common difference = d = 13 – 9 = 4

nth term of AP:

a_{n} = a + (n – 1)d

a_{24} = 9 + (24 – 1)(4)

= 9 + (23)(4)

= 9 + 92

= 101

Therefore, 24th term is 101.

**Question 9: Which term of AP 21, 18, 15,… is -81?**

**Solution:**

Given AP: 21, 18, 15, ….

First term = a = 21

Common difference = d = 18 – 21 = -3

a_{n} = a + (n – 1)d

According to the given,

-81 = 21 + (n – 1)(-3)

-81 – 21 = (n – 1)(-3)

-102 = (n – 1)(-3)

⇒ (n – 1) = (-102)/(-3)

⇒ n – 1 = 34

⇒ n = 34 + 1 = 35

Hence, 35th term of the given AP is -81.

**Question 10: Which term of AP 84, 80, 76,… is zero?**

**Solution:**

Given AP: 84, 80, 76…..

First term = a = 84

Common difference = d = 80 – 84 = -4

a_{n} = a + (n – 1)d

According to the given,

0 = 84 + (n – 1)(-4)

⇒ -84 = (n – 1) × -4

⇒ (n – 1) = (-84)/(-4)

⇒ n – 1 = 21

⇒ n = 21 + 1 = 22

Therefore, 22nd term of the given AP is zero.

**Question 11: Is 301 any term of sequence 5, 11, 17, 23,…?**

**Solution:**

Given AP: 5, 11, 17, 23 …..

First term = a = 5

Common difference = d = 11 – 5 = 6

a_{n} = a + (n – 1)d

According to the given,

301 = 5 + (n – 1)(6)

301 – 5 = 6(n – 1)

⇒ 6(n – 1) = 296

⇒ (n – 1) = 296/6

⇒ n – 1 = 49.33

⇒ n = 49.33 + 1 = 50.33

Value of n cannot be a fraction i.e. it should be a whole number.

Hence, 301 is not the term of given AP.

Question 12: If 6th and 17th terms of an AP are 19 and 41 respectively, then find 40th term.

Solution:

Given,

6th term = a_{6} = 19

17th term = a_{17} = 41

nth term,

a_{n} = a + (n – 1)d

a_{6} = a + (6 – 1)d

19 = a + 5d ….(i)

a_{17} = a + (17 – 1)d

⇒ 41 = a + 16d ….(ii)

Subtracting (i) from (ii),

41 – 19 = a + 16d – a – 5d

22 = -11d

d = -22/11 = -2

Substituting d = -2 in (i),

19 = a + 5(-2)

a – 10 = 19

a = 19 – 10 = 9

a_{40} = a + (40 – 1)d

= 9 + 39 × 2

= 9 + 78

= 87

Therefore, 40th term of the given AP is 87.

**Question 13: Find the 12th term from the last of an AP: 1, 4, 7, 10,…88.**

**Solution:**

Given AP: 1, 4, 7, 10,…, 88

First term = a = 1

Common difference = d = 4 – 1 = 3

Last term = a_{n} = 88

rth term from last = a_{n} – (r – 1)d

12th term from last = 88 – (12 – 1) × 3

= 88 – 11 × 3

= 88 – 33

= 55

Hence, the 12th term from the last term of AP is 55.

**Question 14: There are 60 terms in an AP. If its first and last terms are 7 and 125 respectively, then find its 32th term.**

**Solution:**

Given,

NUmber of terms = n = 60

First term = a = 7

Last term = a_{n} = 125

nth term:

a_{n} = a + (n – 1)d

125 = 7 + (60 – 1)d

125 – 7 = 59d

59d = 118

d = 118/59 = 2

32nd term = a_{32}

= a + (32 – 1)d

= 7 + 31(2)

= 7 + 62

= 69

Therefore, the 32nd term of the AP is 69.

**Question 15: Four numbers are in AP. If the sum of numbers is 50 and the largest number is four times the smaller one, then find the numbers.**

**Solution:**

Let four numbers of an AP be a, a + d, a + 2d, a + 3d.

According to the given,

a + (a + d) + (a + 2d) + (a + 3d) = 50

4a + 6d = 50

2(2a + 3d) = 50

2a + 3d = 25 ….(i)

If larger number is 4 times the smaller number, then equation will be:

a + 3d = 4 x a

a + 3d = 4a

3d = 3a

⇒ d = a ….(ii)

From (i) and (ii),

2a + 3a = 25

5a = 25

a = 25/5 = 5

∴ d = 5 [From (ii)]

Numbers are:

a = 5

a + d = 5 + 5 = 10

a + 2d = 5 + 2 × 5 = 15

a + 3d = 5 + 3 × 5 = 20

Therefore, the four numbers are 5, 10, 15, and 20.

### RBSE Maths Chapter 5: Exercise 5.3 Textbook Important Questions and Solutions

**Question 16: Find the sum of the following arithmetic progression.**

**1, 3, 5, 7,….upto 12 terms.**

**Solution:**

Given AP: 1, 3, 5, 7

First term = a = 1

Common difference = d = 3 – 1 = 2

Sum of first n terms

S_{n} = n/2[2a + (n – 1)d]

S_{12} = (12/2)[2(1) + (12 – 1)2]

= 6 (2 + 22)

= 6 × 24

= 144

Therefore, the sum of the first 12 terms = 144.

**Question 17: Find the sum of the following**

**3 + 11 + 19 + … + 803**

**Solution:**

Given,

3 + 11 + 19 + … + 803

First term = a = 3

Common difference = d = 11 – 3 = 8

nth term of AP = a_{n} = 803

a + (n – 1)d = 803

3 + (n – 1)(8) = 803

(n – 1)(8) = 803 – 3 = 800

n – 1 = 800/8 = 100

n = 100 + 1 = 101

Therefore, the given AP has 101 terms.

Sum of n terms

S_{n} = n/2(a + a_{n})

S_{101} = (101/2)(3 + 803)

= (101 × 806)/2

= 101 × 403

= 40703

Hence, the sum is 40703

**Question 18: How many terms of an AP 63, 60, 57,… taken so that their sum is 693?**

**Solution:**

Given AP: 63, 60, 57,…

First term = a = 63

Common difference = d = 60 – 63 = -3

Let n be the number of terms.

S_{n} = 693

Sum of first n terms

S_{n} = n/2[2a + (n – 1)d]

693 = n/2[2(63) + (n- 1)(-3)]

1386 = n(26 – 3n + 3)

1386 = 129n – 3n^{2}

⇒ 3n2 – 129n + 1386 = 0

⇒ n^{2} – 43n + 462 = 0

⇒ n^{2} – 21n – 22n + 462 = 0

⇒ n(n – 21) – 22(n – 21) = 0

⇒ (n – 21)(n – 22) = 0

⇒ n – 21 = 0 or n – 22 = 0

⇒ n = 21 or n = 22

Therefore, by taking 21 or 22 terms of given AP we will get a sum of 693.

**Question 19: Find the sum of the first 25 terms of the following progression whose nth term is given a _{n} = 7 – 3n.**

**Solution:**

Given,

a_{n} = 7 – 3n

Substituting n = 1, 2, 3,….

a_{1} = 7 – 3(1) = 7 – 3 = 4

a_{2} = 7 – 3(2) = 7 – 6 = 1

a_{3} = 7 – 3(3) = 7 – 9 = -2

a_{2} – a1 = 1 – 4 = -3

a_{3} – a2 = -2 – 1 = -3

Common difference = d = -3

Therefore, AP is 4, 1, -2,…

Here, a = 4, d = -3 and n = 25 (given)

Sum of first n terms is:

S_{n} = n/2[2a + (n – 1)d]

S_{25} = (25/2)[2(4) + (25 – 1)(-3)]

= (25/2)[8 + 24(-3)]

= (25/2)(8 – 72)

= (25/2) × (-64)

= -25 × 32

= -800

Hence, the sum of the first 25 terms of the AP is -800.

**Question 20: Find the sum of all odd numbers divisible by 3 between 1 and 1000.**

**Solution:**

Odd numbers between 1 and 1000, which are divisible by 3 are:

3, 9, 15, 21,…,999

This is an AP with:

First term = a = 3

Common difference = d = 9 – 3 = 6

Let n be the number of terms.

nth term of an AP:

a_{n} = 999

a + (n – 1)d = 999

3 + (n – 1)(6) = 999

(n – 1)(6) = 999 – 3 = 996

n – 1 = 996/6 = 166

n = 166 + 1 = 167

Therefore, number of terms = 167

Sum of n terms

S_{n} = n/2(a + a_{n})

S_{167} = (167/2)(3 + 999)

= (167/2) × 1002

= 167 × 501

= 83667

Hence, the sum of the AP is 83667.

**Question 21: First term of an arithmetic progression is 8, nth term is 33 and sum of first n terms is 123, then find n and common difference d.**

**Solution:**

Given,

First term = a = 8

nth term = a_{n} = 33

Sum of first n terms = S_{n} = 123

S_{n} = n/2(a + a_{n})

123 = n/2(8 + 33)

123 = (n/2) × 41

n/2 = 123/41

n/2 = 3

n = 3 × 2 = 6

nth term:

a_{n} = 33 (given)

a + (n – 1)d = 33

8 + (6 – 1)d = 33

5d = 33 – 8 = 25

d = 25/5 = 5

Therefore, the number of terms (n) = 6 and the common difference (d) = 5.

**Question 22: A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find**

**(i) the production in the 1st year**

**(ii) the production in the 10th year**

**(iii) the total production in the first 7 years.**

**Solution:**

(i) Let the production of TV sets in the first year be a.

And d be the increase in number of TV sets every year.

Given,

Production in the third year = a_{3} = 600

a + (3 – 1)d = 600

a + 2d = 600 ….(i)

Production in the seventh year = a_{7} = 700

a + (7 – 1)d = 700

a + 6d = 700 ….(ii)

Subtracting (i) from (ii),

a + 6d – a – 2d = 700 – 600

4d = 100

d = 100/4 = 25

Substituting d = 25 in (i),

a + 2(25) = 600

a + 50 = 600

a = 600 – 50 = 550

Therefore, the production of TV sets in the 1st year = 550

(ii) Production of TV sets in 10th year

a_{10} = a +(10 – 1)d

= 550 + 9(25)

= 550 + 225

= 775

Therefore, the production of TV sets in the 10th year = 775

(iii) Production of TV sets in 7 years

S_{n} = n/2[2a + (n – 1)d]

S_{7} = (7/2)[2(550) + (7 – 1)(25)]

= (7/2)[1100 + 6(25)]

= (7/2)(1100 + 150)

= (7/2) × 1250

= 7 × 625

= 4375

Therefore, total production of TV sets in 7 years = 4375

### RBSE Maths Chapter 5: Additional Important Questions and Solutions

**Question 1: The common difference between two APs is the same out of two, one AP has first term is 8 and other is 3. The difference between their 30th terms is:**

**(a) 11**

**(b) 3**

**(c) 8**

**(d) 5**

**Solution:**

Correct answer: (d)

30th term of first AP

a_{30} = 8 + (30 – 1)d (first term = 8)

= 8 + 29d

30th term of second AP

b_{30} = 3 + (30 – 1)d (first term = 3)

= 3 + 29d

a_{30} – b_{30} = 8 – 3 = 5

**Question 2: If 18, a, b, -3 are in an AP, then a + b =**

**(a) 19**

**(b) 7**

**(c) 11**

**(d) 15**

**Solution:**

Correct answer: (d)

Given 18, a, b, -3 are in AP.

First term = 18

Second term = 18 + d = a

Third term = 18 + 2d = b

Fourth term = 18 + 3d = -3

⇒ 3d = -3 – 18

⇒ 3d = -21

⇒ d = (-21)/3 = 7

Second term = a = 18 + (-7) = 18 – 7 = 11

Third term = b = 18 + 2 × (-7) = 18 – 14 = 4

Therefore, a + b = 11 + 4 = 15

**Question 3: If 7th and 13th terms of an AP are 34 and 64, respectively, then its 18th term is:**

**(a) 89**

**(b) 88**

**(c) 87**

**(d) 90**

**Solution:**

Correct answer: (a)

Given,

7th term = a7 = 34

a + 6d = 34 ….(i)

13th term = a_{13} = 64

a + 12d = 64 ….(ii)

Subtracting (i) from (ii),

6d = 30

d = 30/6 = 5

Substituting d = 5 in (i),

a + 6(5) = 34

a + 30 = 34

a = 34 – 30 = 4

18th term = a_{18}

= a + 17d

= 4 + 17(5)

= 4 + 85

= 89

**Question 4: If the sum of n terms of an AP is 3n ^{2} + 5n, then which term of series is 164?**

**(a) 12th**

**(b) 15th**

**(c) 27th**

**(d) 20th**

**Solution:**

Correct answer: (c)

Given,

S_{n} = 3n^{2} + 5n

S_{1} = 3(1)^{2} + 5(1) = 3 + 5 = 8

S_{2} = 3(2)^{2} + 5(2) = 12 + 10 = 22

S_{3} = 3(3)^{2} + 5(3) = 27 + 15 = 42

S_{4} = 3(4)^{2} + 5(4) = 48 + 20 = 68

∴ a_{1} = S_{1} = 8

a_{2} = S_{2} – S_{1} = 22 – 8 = 14

a_{3} = S_{3} – S_{2} = 42 – 22 = 20

a_{4} = S_{4} – S_{3} = 68 – 42 = 26

Hence, the AP is 8, 14, 20, 26 …… 164

a = 8,

d = 14 – 8 = 6 and a_{n} = 164

Now,

164 = a + (n – 1)d

164 = 8 + (n – 1)(6)

(n – 1)(6) = 164 – 8 = 156

n – 1 = 156/6 = 26

∴ n = 26 + 1 = 27

Therefore, the 27th term is 164.

**Question 5: If S _{n} is the sum of first n terms of AP and S_{2n} = 3S_{n}, then S_{3n} : S_{n} will be:**

**(a) 10**

**(b) 11**

**(c) 6**

**(d) 4**

**Solution:**

Correct answer: (c)

Given,

S_{2n} = 3S_{n}

(2n/2)[2a + (2n – 1)d] = 3(n/2)[2a + (n – 1)d]

4a + 4nd – 2d = 6a + 3nd – 3d

nd + d = 2a ….(i)

S_{3n} : S_{n} = S_{3n}/S_{n}

= (3n/2)[2a + (3n – 1)d]/ (n/2)[2a + (n – 1)d]

= 3 [nd + d + 3nd – d] / [nd + d + nd – d] (from (i))

= 3 (4nd)/2nd

= 6

**Question 6: The first and last term of AP are 1 and 11, respectively. If sum of its terms is 36, then number of terms will be**

**(a) 5**

**(b) 6**

**(c) 99**

**(d) 11**

**Solution:**

Correct answer: (b)

Given,

First term = a = 1

Last term = l = 11

Sum of first n terms

Sn = n/2(first term + last term)

36 = n/2(1 + 11)

72 = 12n

n = 72/12 = 6

Hence, the number of terms = 6.

**Question 7: Write 5th term from last of AP 3, 5, 7, 9, …, 201.**

**Solution:**

Given AP: 3, 5, 7, 9,…,201

First term = a = 3

Common difference = d = 5 – 3 = 2

Last term = a_{n} = 201

rth term from the last = a_{n} – (r – 1)d

r = 5

5th term from the last = 201 – (5 – 1)(2)

= 201 – (4)(2)

= 201 – 8

= 193

**Question 8: If three consecutive terms of AP are 4/5, a, 2, then find the value of a.**

**Solution:**

Given 4/5, a, 2 are in AP.

Thus, a – (4/5) = 2 – a

a + a = 2 + 4/5

2a = (10 + 4)/5

2a = 14/5

a = (14/5) × (1/2)

a = 7/5

**Question 9: Find the sum of the first 1000 positive integers.**

**Solution:**

First 1000 positive integers are:

1, 2, 3,…, 1000

This is an AP with:

First term = a = 1

Common difference = d = 2 – 1 = 1

Last term = a_{n} = 1000

Sum of first n terms = Sn = n/2[a + a_{n}]

S1000 = (1000/2)[1 + 1000]

= 500 × 1001

= 500500

Therefore, the sum of first 1000 positive integers = 500500

**Question 10: Is 299 be any term in sequence of numbers 5, 11, 17, 23,…?**

**Solution:**

Given AP: 5, 11, 17, 23,…

First term = a = 5

Common difference = d = 11 – 5 = 6

nth term of an AP:

a_{n} = a + (n – 1)d

Let 299 be the nth term.

Thus,

299 = 5 + (n – 1)6

(n – 1)6 = 299 – = 294

n – 1 = 294/6 = 49

n = 49 + 1 = 50

Therefore, the 50th term of the given AP is 299.

**Question 11: Which term of an AP: 20, 19 1/4, 18 1/2, 17 3/4,… is the first negative term?**

**Solution:**

Given AP: 20, 19 1/4, 18 1/2, 17 3/4,…

First term = a = 20

Common difference = d = (19 1/4) – 20 = 77/4 – 20 = (77 – 80)/4 = -3/4

Let the nth term of the given AP be the first negative term.

a_{n} < 0

a + (n – 1)d < 0

20 + (n – 1)(-3/4) < 0

20 – (3n/4) + (3/4) < 0

(83/4) – (3n/4) < 0

83 – 3n < 0

3n > 84

n > 84/3

n > 28

Therefore, the 29th term will be the first negative term of the given AP.

**Question 12: Four numbers are in arithmetic progression. If their sum is 20 and the sum of their squares is 120, then find the numbers.**

**Solution:**

Let the first four terms in an AP be:

a – 3d, a – d, a + d, a + 3d

Given that, the sum of first four terms = 20

a – 3d + a – d + a + d + a + 3d = 20

4a = 20

a = 20/4 = 5

Sum of squares of four numbers = 120

(a – 3d)^{2} + (a – d)^{2} + (a + d)^{2} + (a + 3d)^{2} = 120

a^{2} + 9d^{2} – 6ad + a^{2} + d^{2} – 2ad + a^{2} + d^{2} + 2ad + a^{2} + 9d^{2} + 6ad = 120

4a^{2} + 20d^{2} = 120

4(a^{2} + 5d^{2}) = 120

a^{2} +5d^{2} = 120/4

(5)^{2} + 5d^{2} = 30

5d^{2} = 30 – 25 = 5

d^{2} = 5/5 = 1

d = ±1

When a = 5, d = -1

a – 3d = 5 – 3(-1) = 5 + 3 = 8

a – d = 5 – (-1) = 5 + 1 = 6

a + d = 5 + (-1) = 5 – 1 = 4

a + 3d = 5 + 3(-1) = 5 – 3 = 2

When a = 5, d = 1,

The four terms are 2, 4, 6, 8

Therefore, the first four terms are: 2, 4, 6, 8 or 8, 6, 4, 2.

**Question 13: If the fare of a car for the first kilometer is Rs. 20 and for after 1 kilometer is Rs. 11, then find the total fare for 15 kilometers.**

**Solution:**

Given,

Fare for first kilometer = Rs. 20

Fare for 2 kilometers = Rs. 20 + Rs. 11 = Rs. 31

Fare for 3 kilometers = Rs. 20 + 2(Rs. 11) = Rs. (20 + 22) = Rs. 42

….

The list of fares form an AP with:

First term = a = 20

Common difference = d = 31 – 20 = 11

Total fare for 15 kilometers = a_{15}

= a + (15 – 1)(11)

= 20 + 14 (11)

= 20 + 154

= 164

Hence, required fare is Rs. 164.

**Question 14: Find the sum of all the natural numbers divisible by 5 between 2 and 101.**

**Solution:**

Natural numbers which are divisible by 5 between 2 and 101 are:

5, 10, 15, 20,…, 100

This is an AP with:

First term = a = 5

Common difference = d = 5

Last term (i.e. nth term) = a_{n} = 100

Let n be the number of natural numbers which are divisible by 5 between 2 and 101.

a_{n} = a + (n – 1)d

100 = 5 + (n – 1)5

(n – 1)(5) = 100 – 5 = 95

n – 1 = 95/5 = 19

n = 19 + 1 = 20

Sum of first n terms

S_{n} = n/2(a + an)

S_{20} = 20/2(5 + 100)

= 10 × 105

= 1050

Therefore, the required sum is 1050.

**Question 15: If the second and third terms of an arithmetic progression are 3 and 5 respectively, then find the sum of the first 20 terms of it.**

**Solution:**

Given,

Second term of an AP = a_{2} = 3

Third term = a_{3} = 5

Common difference (d) = a_{3} – a_{2} = 5 – 3 = 2

Thus, first term = a1 (or a) = a_{2} – d = 3 – 2 = 1

Sum of first n terms

S_{n} = n/2[2a + (n – 1)d]

S_{20} = 20/2[2(1) + (20 – 1)2]

= 10[2 + 19(2)]

= 10(2 + 38)

= 10 × 40

= 400

**Question 16: Find the sum of the first 20 terms of the AP: 13, 8, 3,…**

**Solution: **

Given AP:

13, 8, 3,….

First term = a = 13

Common difference = d = 8 – 13 = -5

Sum of first n terms

S_{n} = n/2[2a + (n – 1)d]

S_{20} = 20/2[2(13) + (20 – 1)(-5)]

= 10[26 + 19(-5)]

= 10(26 – 95)

= 10 × (-69)

= -690

**Question 17: Find the 11th term of the AP: -17, -12, -7,…**

**Solution:**

Given AP: -17, -12, -7,….

First term = a = -17

Common difference = d = -12 – (-17) = -12 + 17 = 5

nth term of AP

a_{n} = a + (n – 1)d

a_{11} = -17 + (11 – 1)(5)

= -17 + 10(5)

= -17 + 50

= 33

Therefore, the 11th term of the AP is 33.

**Question 18: Write the common difference of the AP: 7, 5, 3, 1, -1, -3,…**

**Solution:**

Given AP:

7, 5, 3, 1, -1, -3,…

Common difference = Second term – First term

= 5 – 7

= -2

Hence, the common difference = -2