The document NCERT Exemplar Solutions: Arithmetic Progressions Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

All you need of Class 10 at this link: Class 10

**Exercise 5.1**

**Choose the correct answer from the given four options in the following questions:****1. In an AP, if d = –4, n = 7, a _{n }= 4, then a is**

We know that nth term of an AP is

a

where,

a = first term

a

d is the common difference

According to the question,

4 = a + (7 – 1)(- 4)

4 = a – 24

a = 24 + 4 = 28

We know that nth term of an AP is

a

Where,

a = first term

a

d is the common difference

a

= 3.5

(Since, d = 0, it’s a constant A.P)

According to the question,

a

a

a

a

a

a

a

a

Therefore, it’s an A.P with d = 4

(a) –20

(b) 20

(c) –30

(d) 30

First term, a = – 5

Common difference,

d = 5 – (-5/2) = 5/2

n = 11

We know that the nth term of an AP is

a

Where,

a = first term

a

d is the common difference

a

a

(a) – 2, 0, 2, 4

(b) – 2, 4, – 8, 16

(c) – 2, – 4, – 6, – 8

(d) – 2, – 4, – 8, –16

First term, a = – 2

Second Term, d = – 2

a

We know that the nth term of an AP is

a

Where,

a = first term

a

d is the common difference

Hence, we have,

a

Similarly,

a

a

So the A.P is

– 2, – 4, – 6, – 8

(a) 17

(b) 137

(c) 143

(d) –143

First two terms of an AP are a = – 3 and a

We know, nth term of an AP is

a

Where,

a = first term

a

d is the common difference

a

4 = – 3 + d

d = 7

Common difference, d = 7

a

= – 3 + (20)(7)

= 137

**Ans: **(b)**Explanation:**

We know that the nth term of an AP is

a_{n =} a + (n – 1)d

Where,

a = first term

a_{n} is nth term

d is the common difference

a_{2 =} a + d = 13 …..(1)

a_{5 =} a + 4d = 25 …… (2)

From equation (1) we have,

a = 13 – d

Using this in equation (2), we have

13 – d + 4d = 25

13 + 3d = 25

3d = 12

d = 4

a = 13 – 4 = 9

a_{7 =} a + 6d

= 9 + 6(4)

= 9 + 24 = 33**8. Which term of the AP: 21, 42, 63, 84… is 210?****(a) 9 ^{th} **

Let nth term of the given AP be 210.

According to question,

first term, a = 21

common difference, d = 42 – 21 = 21 and a

We know that the nth term of an AP is

a

Where,

a = first term

a

d is the common difference

210 = 21 + (n – 1)21

189 = (n – 1)21

n – 1 = 9

n = 10

So, 10th term of an AP is 210.

Given, the common difference of AP i.e., d = 5

Now,

As we know, nth term of an AP is

a

where a = first term

a

d is the common difference

a

= 5d

= 5(5)

= 25

**Exercise 5.2**

**1. Which of the following form an AP? Justify your answer. (i) –1, –1, –1, –1,…**

We have a

a

a

a

Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.

**(ii) 0, 2, 0, 2,…****Solution:**

We have a_{1} = 0, a_{2} = 2, a_{3} = 0 and a_{4} = 2

a_{2} – a_{1} = 2

a_{3} – a_{2} = – 2

a_{4} – a_{3} = 2

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

**(iii) 1, 1, 2, 2, 3, 3…****Solution:**

We have a_{1} = 1 , a_{2} = 1, a_{3} = 2 and a_{4} = 2

a_{2} – a_{1} = 0

a_{3} – a_{2} = 1

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

**(iv) 11, 22, 33…****Solution:**

We have a_{1} = 11, a_{2} = 22 and a_{3} = 33

a_{2} – a_{1} = 11

a_{3} – a_{2} = 11

Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

**(v) 1/2,1/3,1/4, …****Solution:**

We have a_{1} = ½ , a_{2} = 1/3 and a_{3} = ¼

a_{2} – a_{1} = -1/6

a_{3} – a_{2} = -1/12

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

**(vi) 2, 2 ^{2}, 2^{3}, 2^{4}, …**

We have a

a

a

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

**(vii) √3, √12, √27, √48, …****Solution:**

We have,

a_{1 }= √3, a_{2} = √12, a_{3} = √27 and a_{4} = √48

a_{2} – a_{1 }= √12 – √3 = 2√3 – √3 = √3

a_{3} – a_{2 }= √27 – √12 = 3√3 – 2√3 = √3

a_{4} – a_{3 }= √48 – √27 = 4√3 – 3√3 = √3

Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.**2. Justify whether it is true to say that –1, -3/2, –2, 5/2,… forms an AP as**

**a _{2} – a_{1} = a_{3} – a_{2}.**

False

a

a

a

a

Clearly, the difference of successive terms in not same, all though, a

True

Given

First term, a = – 3

Common difference, d = a

a

= 10d

= – 40

It is so because difference between any two terms of an AP is proportional to common difference of that AP

Suppose there are two AP’s with first terms a and A

And their common differences are d and D respectively

Suppose n be any term

a

A

As common difference is equal for both AP’s

We have D = d

Using this we have

A

= a + (n – 1)d – A – (n – 1)d

= a – A

As a – A is a constant value

Therefore, difference between any corresponding terms will be equal to a – A.

**Exercise 5.3**

**1. Match the APs given in column A with suitable common differences given in column B.**

Column A | Column B |

(A_{1}) 2, – 2, – 6, –10,… | (B_{1}) 2/3 |

(A_{2}) a = –18, n = 10, a= 0_{n} | (B_{2}) – 5 |

(A_{3}) a = 0, a_{10} = 6 | (B_{3}) 4 |

(A_{4}) a_{2 }= 13, a_{4} =3 | (B_{4}) – 4 |

(B_{5}) 2 | |

(B_{6}) 1/2 | |

(B_{7}) 5 |

**Solution:**

(A_{1}) AP is 2, – 2, – 6, – 10, ….

So common difference is simply

a_{2} – a_{1} = – 2 – 2 = – 4 = (B_{3})

(A_{2}) Given

First term, a = – 18

No of terms, n = 10

Last term, a_{n} = 0

By using the nth term formula

a_{n =} a + (n – 1)d

0 = – 18 + (10 – 1)d

18 = 9d

d = 2 = (B_{5})

(A_{3}) Given

First term, a = 0

Tenth term, a_{10} = 6

By using the nth term formula

a_{n =} a + (n – 1)d

a_{10} = a + 9d

6 = 0 + 9d

d = 2/3 = (B_{6})

(A_{4}) Let the first term be a and common difference be d

Given that

a_{2} = 13

a_{4} = 3

a_{2} – a_{4} = 10

a + d – (a + 3d) = 10

d – 3d = 10

– 2d = 10

d = – 5= (B_{1})

**2. Verify that each of the following is an AP, and then write its next three terms.****(i) 0, 1/4, 1/2, 3/4,…****Solution:**

Here,

a_{1 }= 0

a_{2} = ¼

a_{3} = ½

a_{4} = ¾

a_{2} – a_{1} = ¼ – 0 = ¼

a_{3} – a_{2} = ½ – ¼ = ¼

a_{4} – a_{3} = ¾ – ½ = ¼

Since, difference of successive terms are equal,

Hence, 0, 1/4, 1/2, 3/4… is an AP with common difference ¼.

Therefore, the next three term will be,

¾ + ¼ , ¾ + 2(¼), ¾ + 3(¼)

1, 5/4 , 3/2

**(ii) 5, 14/3, 13/3, 4…****Solution:**

Here,

a_{1 }= 5

a_{2} = 14/3

a_{3} = 13/3

a_{4} = 4

a_{2} – a_{1} = 14/3 – 5 = -1/3

a_{3} – a_{2} = 13/3 – 14/3 = -1/3

a_{4} – a_{3} = 4 – 13/3 = -1/3

Since, difference of successive terms are equal,

Hence, 5, 14/3, 13/3, 4… is an AP with common difference -1/3.

Therefore, the next three term will be,

4 + (-1/3), 4 + 2(-1/3), 4 + 3(-1/3)

11/3 , 10/3, 3

**(iii)** √**3 , 2**√**3, 3**√**3,…****Solution:**

Here,

a_{1 }= √3

a_{2} = 2√3

a_{3} = 3√3

a_{4} = 4√3

a_{2} – a_{1} = 2√3 – √3 = √3

a_{3} – a_{2} = 3√3 – 2√3= √3

a_{4} – a_{3} = 4√3 – 3√3= √3

Since, difference of successive terms are equal,

Hence, √3 , 2√3, 3√3,… is an AP with common difference √3.

Therefore, the next three term will be,

4√3 + √3, 4√3 + 2√3, 4√3 + 3√3

5√3, 6√3, 7√3

**(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), …****Solution:**

Here

a_{1} = a + b

a_{2} = (a + 1) + b

a_{3} = (a + 1) + (b + 1)

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

a_{3} – a_{2} = (a + 1) + (b + 1) – (a + 1) – b = 1

Since, difference of successive terms are equal,

Hence, *a *+ *b*, (*a *+ 1) + *b*, (*a *+ 1) + (*b *+ 1), … is an AP with common difference 1.

Therefore, the next three term will be,

(a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1(2), (a + 1) + (b + 1) + 1(3)

(a + 2) + (b + 1), (a + 2) + (b + 2), (a + 3) + (b + 2)

**(v) a, 2a + 1, 3a + 2, 4a + 3,…****Solution:**

Here a_{1} = a

a_{2} = 2a + 1

a_{3} = 3a + 2

a_{4} = 4a + 3

a_{2} – a_{1} = (2a + 1) – (a) = a + 1

a_{3} – a_{2} = (3a + 2) – (2a + 1) = a + 1

a_{4} – a_{3} = (4a + 3) – (3a+2) = a + 1

Since, difference of successive terms are equal,

Hence, *a*, 2*a *+ 1, 3*a *+ 2, 4*a *+ 3,… is an AP with common difference a + 1.

Therefore, the next three term will be,

4a + 3 +(a + 1), 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1)

5a + 4, 6a + 5, 7a + 6**3. Write the first three terms of the APs when a and d are as given below:**

(i) a =1/2, d = -1/6

We know that,

First three terms of AP are:

a, a + d, a + 2d

½, ½ + (-1/6), ½ + 2 (-1/6)

½, 1/3, 1/6

(ii) a = –5, d = –3

We know that,

First three terms of AP are:

a, a + d, a + 2d

-5, – 5 + 1 (- 3), – 5 + 2 (- 3)

– 5, – 8, – 11

(iii) a = √2 , d = 1/√2

We know that,

First three terms of AP are:

a, a + d, a + 2d

√2, √2+1/√2, √2+2/√2

√2, 3/√2, 4/√2

For a, 7, b, 23, c… to be in AP

it has to satisfy the condition,

a

Where d is the common difference

7 – a = b – 7 = 23 – b = c – 23 …(1)

Let us equate,

b – 7 = 23 – b

2b = 30

b = 15 (eqn 1)

And,

7 – a = b – 7

From eqn 1

7 – a = 15 – 7

a = – 1

And,

c – 23 = 23 – b

c – 23 = 23 – 15

c – 23 = 8

c = 31

So a = – 1

b = 15

c = 31

Then, we can say that, the sequence – 1, 7, 15, 23, 31 is an AP

We know that,

The first term of an AP = a

And, the common difference = d.

According to the question,

5

Using the n

a

We get,

a + 4d = 19

a = 19 – 4d …(1)

Also,

13

a + 12d – (a + 7d) = 20

5d = 20

d = 4

Substituting d = 4 in equation 1,

We get,

a = 19 – 4(4)

a = 3

Then, the AP becomes,

3, 3 + 4 , 3 + 2(4),…

3, 7, 11,…

**Exercise 5.4**

**1. The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.****Solution:**

We know that, in an A.P.,

First term = a

Common difference = d

Number of terms of an AP = n

According to the question,

We have,

S_{5} + S_{7} = 167

Using the formula for sum of n terms,

S_{n} = (n/2) [2a + (n-1)d]

So, we get,

(5/2) [2a + (5-1)d] + (7/2)[2a + (7-1)d] = 167

5(2a + 4d) + 7(2a + 6d) = 334

10a + 20d + 14a + 42d = 334

24a + 62d = 334

12a + 31d = 167

12a = 167 – 31d …(1)

We have,

S_{10} = 235

(10/2) [2a + (10-1)d] = 235

5[ 2a + 9d] = 235

2a + 9d = 47

Multiplying L.H.S and R.H.S by 6,

We get,

12a + 54d = 282

From equation (1)

167 – 31d + 54d = 282

23d = 282 – 167

23d = 115

d = 5

Substituting the value of d = 5 in equation (1)

12a = 167 – 31(5)

12a = 167 – 155

12a = 12

a = 1

We know that,

S_{20 }= (n/2) [2a + (20 – 1)d]

= 20/(2[2(1) + 19 (5)])

= 10[ 2 + 95]

= 970

Therefore, the sum of first 20 terms is 970.**2. Find the****(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.****(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.****(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.****[Hint (iii): These numbers will be: multiples of 2 + multiples of 5 – multiples of 2 as well as of 5]****Solution:****(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.**

We know that,

Multiples of 2 as well as of 5 = LCM of (2, 5) = 10

Multiples of 2 as well as of 5 between 1 and 500 = 10, 20, 30…, 490.

Hence,

We can conclude that 10, 20, 30…, 490 is an AP with common difference, d = 10

First term, a = 10

Let the number of terms in this AP = n

Using n^{th} term formula,

a_{n} = a + (n – 1)d

490 = 10 + (n – 1)10

480 = (n – 1)10

n – 1 = 48

n = 49

Sum of an AP,

S_{n} = (n/2) [a + a_{n}], here a_{n} is the last term, which is given]

= (49/2) × [10 + 490]

= (49/2) × [500]

= 49 × 250

= 12250

Therefore, sum of those integers between 1 and 500 which are multiples of 2 as well as of 5 = 12250

**(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.**

We know that,

Multiples of 2 as well as of 5 = LCM of (2, 5) = 10

Multiples of 2 as well as of 5 from 1 and 500 = 10, 20, 30…, 500.

Hence,

We can conclude that 10, 20, 30…, 500 is an AP with common difference, d = 10

First term, a = 10

Let the number of terms in this AP = n

Using n^{th} term formula,

a_{n} = a + (n – 1)d

500 = 10 + (n – 1)10

490 = (n – 1)10

n – 1 = 49

n = 50

Sum of an AP,

S_{n} = (n/2) [ a + a_{n}], here a_{n} is the last term, which is given]

= (50/2) ×[10 + 500]

= 25× [10 + 500]

= 25(510)

= 12750

Therefore, sum of those integers from 1 to 500 which are multiples of 2 as well as of 5= 12750

**(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.**

We know that,

Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2, 5)

Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (10)

Multiples of 2 or 5 from 1 to 500 = List of multiple of 2 from 1 to 500 + List of multiple

of 5 from 1 to 500 – List of multiple of 10 from 1 to 500

= (2, 4, 6… 500) + (5, 10, 15… 500) – (10, 20, 30… 500)

Required sum = sum(2, 4, 6,…, 500) + sum(5, 10, 15,…, 500) – sum(10, 20, 30,., 500)

Consider the first series,

2, 4, 6, …., 500

First term, a = 2

Common difference, d = 2

Let n be no of terms

a_{n} = a + (n – 1)d

500 = 2 + (n – 1)^{2}

498 = (n – 1)^{2}

n – 1 = 249

n = 250

Sum of an AP, S_{n }= (n/2) [ a + a_{n}]

Let the sum of this AP be S_{1,}

S_{1 }= S_{250} = (250/2) ×[2 + 500]

S_{1} = 125(502)

S_{1} = 62750 … (1)

Consider the second series,

5, 10, 15, …., 500

First term, a = 5

Common difference, d = 5

Let n be no of terms

By nth term formula

a_{n} = a + (n – 1)d

500 = 5 + (n – 1)

495 = (n – 1)5

n – 1 = 99

n = 100

Sum of an AP, S_{n }= (n/2) [ a + a_{n}]

Let the sum of this AP be S_{2,}

S_{2 }= S_{100} = (100/2) ×[5 + 500]

S_{2} = 50(505)

S_{2} = 25250 … (2)

Consider the third series,

10, 20, 30, …., 500

First term, a = 10

Common difference, d = 10

Let n be no of terms

a_{n} = a + (n – 1)d

500 = 10 + (n – 1)10

490 = (n – 1)10

n – 1 = 49

n = 50

Sum of an AP, S_{n }= (n/2) [ a + a_{n}]

Let the sum of this AP be S_{3,}

S_{3 }= S_{50} = (50/2) × [2 + 510]

S_{3} = 25(510)

S_{3} = 12750 … (3)

Therefore, the required Sum, S = S_{1} + S_{2} – S_{3}

S = 62750 + 25250 – 12750

= 75250**3. The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.****Solution:**

We know that,

First term of an AP = a

Common difference of AP = d

n^{th} term of an AP, a_{n} = a + (n – 1)d

According to the question,

a_{s} = ½ a_{2}

2a_{8} = a_{2}

2(a + 7d) = a + d

2a + 14d = a + d

a = – 13d …(1)

Also,

a_{11} = 1/3 a_{4} + 1

3(a + 10d) = a + 3d + 3

3a + 30d = a + 3d + 3

2a + 27d = 3

Substituting a = -13d in the equation,

2 (- 13d) + 27d = 3

d = 3

Then,

a = – 13(3)= – 39

Now,

a_{15} = a + 14d

= – 39 + 14(3)

= – 39 + 42

= 3

So 15^{th} term is 3.**4. An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.****Solution:**

We know that,

First term of an AP = a

Common difference of AP = d

n^{th} term of an AP, a_{n} = a + (n – 1)d

Since, n = 37 (odd),

Middle term will be (n+1)/2 = 19^{th }term

Thus, the three middle most terms will be,

18^{th}, 19^{th} and 20^{th} terms

According to the question,

a_{18} + a_{19} + a_{20} = 225

Using a_{n} = a + (n – 1)d

a + 17d + a + 18d + a + 19d = 225

3a + 54d = 225

3a = 225 – 54d

a = 75 – 18d … (1)

Now, we know that last three terms will be 35^{th}, 36^{th} and 37^{th} terms.

According to the question,

a_{35} + a_{36} + a_{37} = 429

a + 34d + a + 35d + a + 36d = 429

3a + 105d = 429

a + 35d = 143

Substituting a = 75 – 18d from equation 1,

75 – 18d + 35d = 143 [ using eqn1]

17d = 68

d = 4

Then,

a = 75 – 18(4)

a = 3

Therefore, the AP is a, a + d, a + 2d….

i.e. 3, 7, 11….**5. Find the sum of the integers between 100 and 200 that are****(i) divisible by 9****(ii) not divisible by 9****[Hint (ii): These numbers will be: Total numbers – Total numbers divisible by 9]****Solution:**

(i) The number between 100 and 200 which is divisible by 9 = 108, 117, 126, …198

Let the number of terms between 100 and 200 which is divisible by 9 = n

a_{n} = a + (n – 1)d

198 = 108 + (n – 1)9

90 = (n – 1)9

n – 1 = 10

n = 11

Sum of an AP = S_{n }= (n/2) [ a + a_{n}]

S_{n} = (11/2) × [108 + 198]

= (11/2) × 306

= 11(153)

= 1683

(ii) Sum of the integers between 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (sum of total numbers between 100 and 200 which is divisible by 9)

Sum, S = S_{1} – S_{2}

Here,

S_{1} = sum of AP 101, 102, 103, – – – , 199

S_{2} = sum of AP 108, 117, 126, – – – , 198

For AP 101, 102, 103, – – – , 199

First term, a = 101

Common difference, d = 199

Number of terms = n

Then,

a_{n} = a + (n – 1)d

199 = 101 + (n – 1)1

98 = (n – 1)

n = 99

Sum of an AP = S_{n }= (n/2) [ a + a_{n}]

Sum of this AP,

S_{1 }= (99/2) × [199 + 101]

= (99/2) × 300

= 99(150)

= 14850

For AP 108, 117, 126, – – – – , 198

First term, a = 108

Common difference, d = 9

Last term, a_{n} = 198

Number of terms = n

Then,

a_{n} = a + (n – 1)d

198 = 108 + (n – 1)9

10 = (n – 1)

n = 11

Sum of an AP = S_{n }= (n/2) [ a + a_{n}]

Sum of this AP,

S_{2 }= (11/2) × [108 + 198]

= (11/2) × (306)

= 11(153)

= 1683

Substituting the value of S_{1} and S_{2} in the equation, S = S_{1} – S_{2}

S = S_{1} + S_{2}

= 14850 – 1683

= 13167

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

62 videos|363 docs|103 tests