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Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 9 |

Chapter Name | Arithmetic Progressions |

Exercise | 9.4 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Solutions for Class 10 Chapter** **9**** Arithmetic Progressions** Ex 9.4 Download PDF

**Arithmetic Progressions**Ex 9.4 Download PDF

**Chapter 9: Arithmetic Progressions Exercise 9.4**

**Question: 1**

If 12^{th} of an A.P is 82 and 18^{th} term is 124. Then find out the 24^{th} term.

**Solution:**

Given: a_{12} = 82 and a_{18} = 124

we know :

a_{n} = a + (n – 1) c. d

⟹ a_{12} = a + (12 – 1) c. d

⟹ 82 = a + 11 c. d … (1)

⟹ 124 = a + (18 – 1) c. d

⟹ 124 = a + 17 c. d …. (2)

Subtracting (2) from (1)

⟹ (a + 17 c. d) – (a + 11c.d) = 124 – 82

⟹ a +17c.d – a – 11c.d = 42

⟹ 6 c.d = 42

⟹ c. d = 7

Here we have,

Common Difference (c. d) = 7 putting c. d = 7 in equation (1), we get

⟹ a + 11 × 7 = 82

⟹ a = 82 – 77

⟹ a = 5

Now, we have

First Term (a) = 5 we have to find 24^{th} term

a_{24} = a + (24 – 1) c. d

= 5 + 23 × 7

= 5 + 161 = 166

**Question: 2**

In an A.P. the 24^{th} term is twice the 10^{th }term. Prove that the 72^{nd} term is twice the 34^{th} term.

**Solution:**

Given

24^{th} term is twice the 10^{th} term a_{24} = 2× a_{10 } . . . . . . (1)

Let, first term be a and common difference be

d we know, n^{th} term is a_{n} = a + (n – 1)d

from equation (1), we have

a + (24 – 1)d = 2(a + (10 – 1)d)

⟹ a + 23d = 2 (a + 9d)

⟹ a + 23 d = 2a + 18d

⟹ (23 – 18)d = a

⟹ a = 5d

We have to prove that, 72^{nd} term is twice the 34^{th} term

⟹ a_{72} = 2 × a_{34}

⟹ a + (72 – 1)d = 2 [a + (34 – 1)d]

⟹ a + 71d = 2(a + 33d)

⟹ a + 71d = 2a + 66d

Putting the value of a = 5 in the above equation,

⟹ 5d + 71d = 2 (5d) + 66d

⟹ 76d = 76d Hence it is proved…

**Question: 3**

If the (m + 1)^{th} term of an A.P. is twice the (n + 1)^{th} term of the A.P. Then prove that: (3m + 1)^{th }is twice the (m + n + 1)^{th} term.

**Solution:**

From the question, we have

a_{(m+1)} = 2 a_{(n+1)}

Let, First term = a and Common Difference = d

⟹ a + (m + 1 – 1)d = 2[a+(n + 1 – 1)d]

⟹ a + md = 2a + 2nd

⟹ a = md – 2nd

⟹ a = (m – 2n)d …. (2)

We have to prove, a_{(3m+1)} = 2a_{(m+n+1)}

⟹ a + (3m + 1 – 1)d = 2 [a + (m + n + 1 -1)d]

⟹ a + 3md = 2a + 2(m + n)d putting the value of a = (m – 2n)d, from equation (1)

⟹ (m – 2n)d + 3md = 2[(m – 2n)d] + 2( m + n)d

⟹ m – 2n + 3m = 2m – 4n + 2m + 2n

⟹ 4m – 2n = 4m – 2n Hence it is proved…

**Question: 4**

If the n^{th} term of the A.P. 9, 7, 5, … is same as the n^{th} term of the A.P. 15, 12 , 9, … find n.

**Solution:**

We have here, First sequence is 9, 7, 5, … First term (a) = 9, Common Difference (c. d) = 9 – 7 = – 2 n^{th} term

= a + (n-1) c. d

⟹ a_{n} = 9 + (n – 1)(– 2)

= 9 – 2n + 2 = 11 – 2n

Second sequence is 15, 12, 9, … here,

First term (a) = 15

Common Difference (c. d) = 12 -15 = -3 n^{th} term = a + (n -1)d

⟹ a’_{n} = 15 + (n – 1)(-3)

= 15 – 3n + 3 = 18 – 3n

We are given in the question that the n^{th} term of both the A.P.s are same, So, we can write it as

a^{}_{n }= a’_{n}

⟹ 11 – 2n = 18 – 3n

⟹ **n = 7**

So, the 7^{th} term of both the A.P.s will be equal.

**Question: 5**

Find the 13^{th} term from the end in the following A.P.

(i) 4, 9, 14, … , 254.

(ii) 3, 5, 7, 9, …, 201.

**(iii) 1, 4, 7, 10, … , 88.**

**Solution:**

(i) We have,

First term (a) = 4 and common difference (c. d)

= 9 – 4 = 5

last term here (l) = 254 n^{th} term from the end is : l – (n – 1)d we have to find 13^{th} term from end then :

l – 12d = 254 – 12 × 5

= 254 – 60 = 194

(ii). 3, 5, 7, 9, …, 201.

**Solution: **

we have, First term (a) = 3 and common difference (c. d)

= 5 – 3 = 2

last term here (l) = 201 n^{th} term from the end is :

l – (n-1)d we have to find 13^{th} term from end then :

l – 12d = 201 – 12 × 2

= 201 – 24 = 177

(iii). 1, 4, 7, 10, … , 88.

We have, First term (a) = 1 and common difference (c. d)

= 4 -1 = 3 last term here (l) = 88 n^{th} term from the end is:

l – (n-1)d we have to find 13^{th} term from end then:

l – 12d = 88 – 12 × 3

= 88 – 36 = 52

**Question: 6**

The 4^{th} term of an A.P. is three times the first term and the 7^{th} term exceeds the twice the third term by 1. Find the A.P.

**Solution:**

Given, 4^{th} term of the A.P = thrice the first term

⟹ a_{4} = 3

first term Assuming first term to be ‘a’ and the common difference be ‘d’ we have,

a+ ( 4 -1 )d = 3

a ⟹ a + 3d = 3

a ⟹ a = 32d … (1) and also it is given that, the 7^{th} term exceeds the twice of the 3^{rd} term by

1 ⟹ a_{7} + 1 = 2 × a_{3}

⟹ a + (7 – 1)d + 1 = 2[a + (3 – 1)d]

⟹ a + 6d + 1 = 2a + 4d

⟹ a = 2d + 1 … (2)

Putting the value of a = 32d from equation (1) in equation (2)

32d = 2d + 1

⟹ 32d – 2d = 1

⟹ 3d − 4d2 = 1

⟹ – d = 2

⟹ d = – 2 put d = -2 in a = 32d

⟹ a = 32(-2)

⟹ a = – 3

Now, we have a = – 3 and d = – 2, so the A.P. is -2, – 5, – 8, -11, …

**Question: 7**

Calculate the third term and the nth term of an A.P. whose 8^{th} term and 13^{th} term are 48 and 78 respectively.

**Solution:**

Given, a_{8} = 48 and a_{13} = 78 n^{th} term of an A.P. is:

a_{n} = a + (n -1)d

so, a_{8} = a + (8 – 1)d = a + 7d …. (1)

a_{13} = a + (13 – 1)d = a + 12d …. (2)

Equating (1) and (2), we get.

⟹ a + 12d – (a + 7d) = 78 – 48

⟹ a + 12d – a -7d = 30

⟹ 5d = 30

⟹ d = 6

Putting the value of d = 6 in equation (1),

a + 7 × 6 = 48

⟹ a + 42 = 48

⟹** **a = 4

Now, we have the first term (a) and the common difference (d) with us, So, n^{th} term will be:

a_{n =} a + (n-1)d

= 4 + (n-1)6 = 4 + 6n – 6

a_{n }= 6n – 2 and the 3^{rd} term will be

a_{3} = 6 × 3 – 2

a_{n} = 16

**Question: 8**

How many three digit numbers are divisible with 3?

**Solution:**

We know the first three digit number which is divisible by 3 is 102 and the last three digit number which is divisible by 3 is 999 So, here we have First term (a) = 102 Common Difference (c. d) = 3 last term or n^{th} term (l) = 999

⟹ a_{n }= 999

⟹ a + (n – 1)c. d = 999

⟹ 102 + (n – 1)3 = 999

⟹ 102 + 3n – 3 = 999

⟹ 99 + 3n = 999

⟹ 3n = 900

⟹ n = 300

Therefore, there are 300 terms in the sequence.

**Question: 9**

An A.P. has 50 terms and the first term is 8 and the last term is 155. Find the 41stterm from the A.P.

**Solution:**

Given, First term (a) = 8 Number of terms (n) = 50 Last term (a_{n}) = 148

⟹ a_{n }= a + (n – 1)d

⟹ 155 = 8 + (50 – 1)d

⟹ 49 d = 147

⟹ d = 3 now, 41^{st} term will be:

a + (41 – 1)d

⟹ 8 + 40 × 3

⟹ 128

**Question: 10**

The sum of 4^{th} and 8th term of an A.P. is 24 and the sum of the 6th and 10^{th} terms is 34. Find the first term and the common difference of the A.P.

**Solution:**

Let’s assume first term be a and common difference be d

Given 4^{th} term + 8^{th} term = 24

⟹ a_{4} + a_{8} = 24

⟹ (a + (4 – 1)d) + (a + (8 – 1)d) = 24

⟹ a + 3d + a + 7d = 24

⟹ 2a + 10d = 24 …. (1)

And 6^{th} term + 10^{th} term = 34

⟹ a_{6 }+ a_{10} = 34

⟹ (a + 5d) + (a + 9d) = 34

⟹ 2a + 14d = 34 …. (2)

Subtracting equation (1) from (2), We get

⟹ (2a + 14d) – (2a + 10d) = 34 -24

⟹ 2a + 14d – 2a – 10d = 10

⟹ 4d = 10

⟹ d = 52 Put d = 52 in equation (1)

⟹ 2a + 10 × 52 = 24

⟹ 2a + 25 = 24

⟹ 2a = -1

⟹ a = −12

Therefore, we have a = −12 and d = 52

**Question: 11**

The first term of an A.P. is 7 and its 100^{th} term is – 488, Find the 50^{th} term of the same A.P.

**Solution:**

Given, First term

(a) = 7 100^{th} term (a_{100}) = -488

We know,

a_{n} = a + (n – 1)d

⟹ (a_{100}) = a + (100 -1)d

⟹ 7 + 99d = – 488

⟹ 99d = – 495

⟹** **d = – 5

Now, we have the common difference (d) = 5

We have to find out the 50^{th} term of the A.P.

Then,

a_{50} = a + 49 d = 7 + 49 × (-5)

= 7 – 245 =** – **238** **

So, the 50^{th} term of the A.P. is – 238

**Question: 12**

Find a_{40} – a_{30} of the following A.P.

(i) 3, 5, 7, 9, . . .

(ii) 4, 9, 14, 19, . . .

**Solution:**

(i) Provided A.P. is 3, 5, 7, 9, . . . So, we have first term (a) = 3 and the common difference (d) is 5 – 3 = 2 we have to find

a_{40 }– a_{30 }= (a + 39d) – (a + 29d)

= a + 39d – a – 29d = 10d

= 10 × 2 =** **20

(ii) Given A.P. is 4, 9, 14, 19, . . .

Common difference (d) = 9 – 4 = 5 we have to find

a_{40 }– a_{30 }= 10d = 10 × 5 = 50

**Question: 13**

Write the expression a_{m} – a_{n} for the A.P. a, a + d, a + 2d, . . .

**Solution:**

General Arithmetic Progression a, a + d, a + 2d, . . . a_{m} – a_{n} = (a + (m – 1)d) – ( a + ( n – 1 )d)

⟹ a + md – d – a – nd + d

⟹ md – kd

⟹ (m – n)d …. (1)

Hence find the common difference of the A.P. for which

(i) 11^{th} term is 5 and 13^{th} term is 79

Given, 11^{th} term (a_{11}) = 5 and 13^{th} term (a_{13}) = 79 from equation (1), taking m = 11 and n = 13

⟹ a_{m} – a_{n }= (13 – 11)d

⟹ 79 – 5 = 2d

⟹ 74 = 2d

⟹ d = 37

(ii) a_{10} – a_{5} = 200

Given, here we have the difference between the 10^{th} term and 5^{th} term.

Putting the value of m and n in equation (1) as 10 and 5, we have

⟹ a_{10} – a_{5} = (10 – 5 )d

⟹ 200 = 5 d

⟹** **d = 40

(iii) 20^{th} term is 10 more than the 18^{th} term

Given,

a_{20} + 10 = a_{18}

⟹ a_{20} – a_{18} = 10 from equation (1), we have

a_{m} – a_{n }= (m – n)d

⟹ a_{20} – a_{18 }= (20 – 18) d

⟹ 10 = 2d

⟹ d = 5

**Question: 14**

Find n if the given value of x is the n term if the given A.P.

(iii) – 1, – 3, – 5, – 7, . . . : x = – 151

(iv) 25, 50, 70, 100, . . . : x = 1000

**Solution:**

(i) Given sequence is

first term (a) = 1

Common difference (d)

n^{th} term a_{n} = a + (n-1) × d

(ii)** **Given sequence is

(iii)** **Given sequence is, -1 , -3, -5, -7, . . . 😡 = -151 first term (a) = -1

Common Difference (d) = -3 – (-1) = -3 + 1 = -2 n^{th} term

a_{n} = a + (n – 1) × d

⟹ -151 = -1 + (n – 1) X – 2

⟹ -151 = -1 – 2n + 2

⟹ -151 = 1 – 2n

⟹ 2n = 152

⟹ n = 76

(iv) Given sequence is, 25, 50, 70, 100, . . . 😡 = 1000

First term (a) = 25

Common Difference (d) = 50 – 25 = 25 n^{th} term a_{n} = a + (n – 1) × d we have

a_{n} = 1000

⟹ 1000 = 25 + ( n – 1 ) 25

⟹ 975 = ( n – 1)25

⟹ n – 1 = 39

⟹ n = 40

**Question: 15**

If an A.P. consists of n terms with the first term a and n^{th} term 1. Show that the sum of the m^{th} term from the beginning and the m^{th} term from the end is (a + 1).

**Solution:**

First term of the sequence is a Last term (l) = a + (n – 1) d

Total no. of terms = n

Common Difference = d m^{th} term from the beginning

a_{m} = a + ( n – 1 )d m^{th} term from the end

= l + (n – 1)(-d)

⟹ a_{(n – m + 1)} = l – (n – 1)d

⟹ a_{m} + a_{(n – m + 1)}

= a + (n – 1)d + (l – ( n – 1)d)

= a + (n-1)d + l – (n – 1)d = a + l

**Question: 16**

Find the A.P. whose third term is 16 and seventh term exceeds its fifth term by 12.

**Solution:**

Given, a_{3} = 16

⟹ a + (3 – 1)d = 16

⟹ a + 2d = 16 … (i) and a_{7} – 12 = a_{5}

⟹ a + 6d -12 = a + 4d

⟹ 2d = 12

⟹ d = 6 Put d = 6 in equation (1) a + 2 X 6 = 16

⟹ a + 12 = 16

⟹ a = 4.

So, the sequence is 4, 10, 16, . . .

**Question: 17**

The 7^{th} term of an A.P is 32 and its 13^{th} term is 62. Find the A.P.

**Solution:**

Given a_{7} = 32

⟹ a + (7 – 1)d = 32

⟹ a + 6d = 32 … (i) and a_{13} = 62

⟹ a + (13 – 1)d = 62

⟹ a + 12d = 62 … (ii) equation (ii) – (i), we have (a + 12d) – (a + 6d) = 62 – 32

⟹ 6d = 30

⟹ d = 5 Putting d = 5 in equation (i) a + 6 × 5 = 32

⟹ a = 32 – 30

⟹ a = 2

So, the obtained A.P. is 2, 7, 12, 17, . . .

**Question: 18**

Which term of the A.P. 3, 10, 17, . . . will be 84 more than its 13^{th} term?

**Solution:**

Given A.P. is 3, 10 , 17, . . . First term (a) = 3

Common Difference (d) = 10 -3 = 7

Let n^{th} term of the A.P. will be 84 more than its 13^{th} term, then a_{n} = 84 + a_{13}

⟹ a + (n – 1)d = a + (13 – 1)d + 84

⟹ (n – 1) × 7 = 12 × 7 + 84

⟹ n – 1 = 24

⟹ n = 25

Hence, 25^{th} term, of the given A.P. is 84 more than the 13^{th} term.

**Question: 19**

Two arithmetic progressions have the same common difference. The difference between their 100^{th} terms is 100. What is the difference between their 1000^{th} terms?

**Solution:**

Let the two A.P. be a_{1}, a_{2} , a_{3} , . . . and b_{1}, b_{2} , b_{3}, . . . a_{n} = a1 + (n – 1)d and b_{n} = a1 + (n – 1)d

Since common difference of two equation is same and given difference between 100^{th} terms is 100

⟹ a_{100} – b_{100} = 100

⟹ a + (100 – 1)d – [b + (100 – 1)d] = 100

⟹ a + 99d -b – 99d = 100

⟹ a + b = 100 … (1)

Difference between 100^{th} term is

⟹ a_{1000} – b_{1000} = a + (1000 – 1)d – [b + (1000 – 1)d]

= a + 999d – b – 999d = a – b = 100 (from equation 1)

Therefore, Difference between 1000^{th} terms of two A.P. is 100.

**Question: 20**

For what value of n, the n^{th} terms of the Arithmetic Progression 63, 65, 67, . . . and , 3, 10, 17, . . . are equal?

**Solution:**

Given two A.P.s are: 63, 65, 67, . . . and 3, 10, 17, . . .

First term for first A.P. is (a) = 63

Common difference (d) is 65 – 63 = 2 n^{th} term (a_{n}) = a + (n – 1)d = 63 + (n – 1)2

First term for second A.P. is (a’) = 3

Common Difference (d’) = 10 – 3 = 7 n^{th} term (a’_{n}) = a’ + (n – 1)d = 3 + (n – 1) 7

Let n^{th } term of the two sequence be equal then, ⟹ 63 + (n – 1)2 = 3 + (n – 1)7

⟹ 60 = ( n – 1 ).7 – ( n – 1 ).2

⟹ 60 = 5(n – 1)

⟹ n – 1 = 12

⟹ n = 13

Hence, the 13^{th} term of both the A.P.s are same.

**Question: 21**

How many multiple of 4 lie between 10 and 250?

**Solution:**

Multiple of 4 after 10 is 12 and multiple of 4 before 250 is 120/4, remainder is 2, so, 250 – 2 = 248

248 is the last multiple of 4 before 250 the sequence is 12,. . . . . . . . , 248 with first term (a) = 12

Last term (l) = 258

Common Difference (d) = 4 n^{th} term (a_{n}) = a + (n – 1)d

Here n^{th} term a _{n} = 248

⟹ 248 = a + ( n – 1 )d

⟹ 12 + (n – 1)4 = 248

⟹ (n – 1)4 = 236

⟹ n – 1 = 59

⟹ n = 59 + 1

⟹ N = 60

Therefore, there are 60 terms between 10 and 250 which are multiples of 4

**Question: 22**

How many three digit numbers are divisible by 7?

**Solution:**

The three digit numbers are 100, . . . . . . , 999 105 us the first 3 digit number which is divisible by 7 and when we divide 999 with 7 remainder is 5, so, 999 – 5 = 994

994 Is the last three digit number which is divisible by 7.

The sequence here is 105, .. . . . . . . . , 994

First term (a) = 105 Last term (l) = 994

Common Difference (d) = 7

Let there are n numbers in the sequence then,

⟹ a_{n} = 994

⟹ a + (n – 1)d = 994

⟹ 105 + (n – 1)7 = 994

⟹ (n – 1) × 7 = 889

⟹ n – 1 = 127

⟹ n = 128

Therefore, there are 128 three digit numbers which are divisible by 7.

**Question: 23**

Which term of the A.P. 8, 14, 20, 26, . . . will be 72 more than its 41^{st} term?

**Solution:**

Given sequence 8, 14, 20, 26, . . . Let its n term be 72 more than its 41^{st} term

⟹ a_{n} = a_{41} + 72 … (1)

For the given sequence, first term (a) = 8,

Common Difference (d) = 14 – 8 = 6 from equation (1), we have

a_{n} = a_{41} + 72

⟹ a + (n – 1)d = a + (41 – 1)d + 72

⟹ 8 + (n – 1)6 = 8 + 40 × 6 + 72

⟹ (n – 1)6 = 312

⟹ n – 1 = 52

⟹ n = 53

Therefore, 53^{rd }term is 72 more than its 41^{st} term.

**Question: 24**

Find the term of the Arithmetic Progression 9, 12, 15, 18, . . . which is 39 more than its 36^{th} term.

**Solution:**

Given A.P. is 9, 12, 15, 18 , . . .

Here we have, First term (a) = 9

Common Difference (d) = 12 – 9 = 3

Let its nth term is 39 more than its 36th term So,

a_{n} = 39 + a_{36}

⟹ a + (n – 1)d = 39 + a + (36 – 1)d

⟹ (n – 1)3 = 39 + 35 × 3

⟹ (n – 1)3 = (13 + 35)3

⟹ n – 1 = 48

⟹ n = 49

Therefore, 49^{th} term of the A.P. 39 more than its 36^{th} term.

**Question: 25**

Find the 8^{th} term from the end of the A.P. 7, 10, 13, . . . , 184.

**Solution:**

Given A.P. is 7, 10, 13, . . . , 184

First term (a) = 7

Common Difference (d) = 10 – 7 = 3 last term (l) = 184 n^{th} term from end = l – ( n – 1 )d

8^{th} term from end = 184 – ( 8 – 1 )3 =

184 – 7 X 3 =

184 – 21 = 183

Therefore, 8^{th} term from the end is 183

**Question: 26**

Find the 10^{th} term from the end of the A.P. 8, 10, 12, . . . , 126

**Solution:**

Given A.P. is 8, 10, 12, . . . , 126

First term (a) = 8

Common Difference (d) = 10 – 8 = 2

Last term (l) = 126 n^{th} term from end is: l – (n – 1)d

So, 10^{th} term from end is: l – (10 – 1)d

= 126 – 9 × 2 = 126 – 18 = 108

Therefore, 109 is the 10^{th} term from the last in the A.P. 8, 10, 12, . . 126.

**Question: 27**

The sum of 4^{th} and 8^{th} term of an A.P. is 24 and the sum of 6^{th} and 10^{th} term is 44. Find the Arithmetic Progression.

**Solution:**

Given a_{4} + a_{8} = 24

⟹ a + (4 – 1)d + a + (8 – 1)d = 24

⟹ 2a + 3d + 7d = 24

⟹ 2a + 10d = 24 … (1) and a_{6} + a_{10} = 44

⟹ a + (6 – 1)d + a + (10 – 1)d = 44

⟹ 2a + 5d + 9 d = 44

⟹ 2a + 14d = 44 … (2) equation (2) – equation (1), we get

2a + 14d – (2a + 10d) = 44 – 24

⟹ 4d = 20

⟹ d = 5 Put d = 5 in equation (1), we get

2a + 10 × 5 = 24

⟹ 2a = 24 – 50

⟹ 2a = – 26

⟹ a = -13 The A.P is -13, – 7, – 2, . . .

**Question: 28**

Which term of the A.P. is 3, 15, 27, 39, . . . will be 120 more than its 21st term?

**Solution:**

Given A.P. is 3, 15, 27, 39, . . . First term (a) = 3

Common Difference (d) = 15 – 3 = 12

Let n^{th} term is 120 more than 21^{st} term

⟹ a_{n} = 120 + a_{21}

⟹ a + (n – 1)d = 120 + a + ( 21 – 1 )d

⟹ (n – 1)d = 120 + 20d

⟹ (n – 1)12 = 120 + 20 × 12

⟹ n – 1 = 10 + 20

⟹ n = 31

Therefore, 31^{st} term of the A.P. is 120 more than the 21^{st} term.

**Question: 29**

The 17^{th} term of an A.P. is 5 more than twice its 8^{th} term. If the 11^{th} term of the A.P. is 43. Find the n^{th} term.

**Solution:**

Given 17^{th} term of an A.P is 5 more than twice its 8^{th} term

⟹ a_{17} = 5 + 2a_{8}

⟹ a + (17 – 1)d = 5 + 2 [a + (8 – 1)d]

⟹ a + 16d = 5 + 2a + 14d

⟹ a + 5 = 2d … (1) and 11^{th} term of the A > P. is 43 a_{11} = 43

⟹ a + (11 – 1)d = 43

⟹ a + 10d = 43

⟹ a + 5 × 2d = 43 from equation (1)

⟹ a + 5 × (a + 5) = 43

⟹ a + 5a + 25 = 43

⟹ 6a = 18

⟹ a = 3

Putting the value of a = 3, in equation (1), we get 3 + 5 = 2d

⟹ 2d = 8

⟹ d = 4

We have to find the n^{th} term (a_{n}) = a + (n – 1)d

= 3 + (n – 1)4 = 3 + 4n – 4 = 4n – 1

Therefore, n^{th} term is 4n – 1

**Question: 30**

Find the number of all three digit natural number which are divisible by 9.

**Solution:**

First three-digit number that is divisible by 9 is 108. Next number is 108 + 9 = 117.

And the last three-digit number that is divisible by 9 is 999.

Thus, the progression will be 108, 117, …. , 999.

All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term

(a): 108 last term (l) = 999 and the common difference (d) as 9 We know that, n^{th} term

(a_{n}) = a + (n – 1)d

According to the question,

999 = 108 + (n – 1)9

⟹ 999 = 108 + 9n – 9

⟹ 999 = 99 + 9n

⟹ 999 = 9n

⟹ 999 – 99

⟹ 9n = 900

⟹ n = 100

Therefore, There are 100 three digit terms which are divisible by 9.

**Question: 31**

The 19^{th} term of an A.P. is equal to three times its 6^{th} term. if its 9^{th} term is 19, find the A.P

**Solution:**

Let a be the first term and d be the common difference.

We know that, nth term

= an= a + (n – 1)d

According to the question, a_{19} = 3a_{6}

⟹ a + (19 – 1)d = 3(a + (6 – 1)d)

⟹ a + 18d = 3a + 15d

⟹ 18d- 15d = 3a – a

⟹ 3d = 2a

⟹ a = 32d …. (1) Also, a9 = 19

⟹ a+(9 – 1)d= 19

⟹ a + 8d = 19 … (2)

On substituting the values of (1) in (2), we get

⟹ 32d + 8d = 19

⟹ 3d + 16d = 19 × 2

⟹ 19d = 38

⟹ d = 2 Now, a = 32 × 2 [From (1)] a = 3 Therefore, The A.P. is : 3, 5, 7, 9, . . .

**Question: 32**

The 9^{th} term of an A.P. is equal to 6 times its second term. If its 5^{th} term is 22, find the A.P.

**Solution:**

Let a be the first term and d be the common difference.

We know that, nth term (a_{n}) = a + (n – 1)d

According to the question, a9 = 6a2

⟹ a + (9 – 1)d = 6(a + (2 – 1)d)

⟹ a + 8d = 6a + 6d

⟹ 8d – 6d = 6a – a

⟹ 2d = 5a

⟹ a = 25 …. (1) Also, a_{5} = 22

⟹ a + (6 – 1)d = 22

⟹ a + 4d = 22 … (2)

On substituting the values of (1) in (2), we get

2/5 d + 4d = 22

⟹ 2d + 20d = 22 × 5

⟹ 22d = 110

⟹ d = 5 Now, a = 25 × 5 [From (1)] ⟹ a = 2

Thus, the A.P. is : 2, 7, 12, 17, . . .

**Question: 33**

The 24^{th} term of an A.P. is twice its 10^{th} term. Show that its 72^{nd} term is 4 times its 15^{th} term.

**Solution:**

Let a be the first term and d be the common difference.

We know that, n^{th} term

(a_{n}) = a + (n – 1)d

According to the question, a_{24} = 2

a_{10} ⟹ a + (24 – 1)d = 2(a + (10 – 1)d)

⟹ a + 23d = 2a + 18d

⟹ 23d – 18d = 2a – a

⟹ 5d = a

⟹ a = 5d …. (1)

Also, a_{72} = a + (72 – 1)

d = 5d + 71d [From (1)] = 6d ….(2) and a_{15} = a + (15 – 1) d = 5d + 14d [From (1)] = 19d …. (3)

On comparing (2) and (3), we get

⟹ 76d = 4 × 19 d

⟹ a_{72} = 4 × a_{15}

Thus, 72^{nd} term of the given A.P. is 4 times its 15^{th} term.

**Question: 34**

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

**Solution:**

Since, the number is divisible by both 2 and 5, means it must be divisible by 10.

In the given numbers, first number that is divisible by 10 is 110.

Next number is 110 + 10 = 120.

The last number that is divisible by 10 is 990.

Thus, the progression will be 110, 120, …, 990.

All the terms are divisible by 10, and thus forms an A.P. having first term as 110 and the common difference as 10.

We know that, n^{th} term = a_{n}= a + (n – 1)d

According to the question, 990 = 110 + ( n – 1 )10

⟹ 990 = 110 + 10n – 10

⟹ 10n = 990 – 100

⟹ n = 89

Thus, the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 is 89.

**Question: 35**

If the7^{th} term of an A.P. is 1/9 and its 9^{th} term is 1/7, find the 63^{rd} term.

**Solution:**

Let a be the first term and d be the common difference.

We know that, nth term = an= a + (n – 1)d

According to the question, a_{7} = 1/9

⟹ a+(7 – 1)d = 1/9

⟹ a+ 6d = 1/9 ….(1) Also, a_{9} = 1/7

⟹ a + (9 – 1)d = 1/7

⟹ a + 8d = 1/7 ….(2)

On Subtracting (1) from (2), we get

⟹ 8d – 6d = 1/7 – 1/9

⟹2d = (9 – 7)/63

⟹ 2d = 2/63

⟹ d = 1/63

Put value of d = 1/63 in equation (1), we get

⟹ a + 6 × 1/63 = 1/9

⟹ a = 1/9 – 6/63

⟹a = (7 – 6)/63

⟹ a = 1/63

Therefore, a_{63} = a + (63 – 1)d = 1/63 + 62/63 = 63/63 = 1

Thus, 63^{rd} term of the given A.P. is 1.

**Question: 36**

The sum of 5^{th} and 9^{th} terms of an A.P. is 30. If its 25^{th} term is three times its 8^{th} term, Find the A.P.

**Solution:**

Let a be the first term and d be the common difference.

We know that, nth term (a_{n}) = a + (n – 1)d

According to the question, a_{5} + a_{9} = 30

⟹ a + (5 – 1)d + a + (9 – 1)d = 30

⟹ a + 4d + a + 8d = 30

⟹ 2a + 12d = 30

⟹ a + 6d = 15 …. (1) Also, a_{25} = 3(a_{8})

⟹ a + (25 – 1)d = 3[a + (8 – 1)d]

⟹ a + 24d = 3a + 21d

⟹ 3a – a = 24d – 21d

⟹ 2a = 3d

⟹ a = 32d ….(2)

Substituting the value of (2) in (1), we get 32d + 6d = 15

⟹ 32d + 6d = 15

⟹ 3d + 12d = 15 × 2

⟹ 15d = 30

⟹ d = 2 now, a = 32d × 2 [From (1)] ⟹ a = 3 Therefore, the A.P. is 3, 5, 7, 9, . . .

**Question: 37**

Find whether 0 (zero) is a term of the A.P. 40, 37, 34, 31, . . .

**Solution:**

Let a be the first term and d be the common difference.

We know that, nth term = an = a + (n – 1)d

It is given that a = 40, d = -3 and a_{n}= 0

According to the question, ⟹ 0 = 40 + (n – 1)(-3)

⟹ 0 = 40 – 3n + 3

⟹ 3n = 43

⟹ n = 433 …. (1)

Here, n is the number of terms, so must be an integer.

Thus, there is no term where 0 (zero) is a term of the A.P. 40, 37, 34, 31,. . .

**Question: 38**

Find the middle term of the A.P. 213, 205, 197, . . . 37.

**Solution:**

Let a be the first term and d be the common difference. We know that, nth term (a_{n}) = a + (n – 1)d

It is given that a = 213, d = – 8 and a_{n} = 37

According to the question, ⟹ 37 = 213 + (n – 1)(-8)

⟹ 37 = 213 – 8n + 8

⟹ 8n = 221 – 37

⟹ an = 184

⟹ n = 23 …. (1)

Therefore, total number of terms is 23. Since, there is odd number of terms.

So, Middle term will be 23 + 12th term, i.e., the 12th term.

a_{12 }= 213 + (12 – 1)(-8)

a_{12 }= 213 – 88 = 125

Thus, the middle term of the A.P. 213, 205, 197, . . . , 37 is 125.

**Question: 39**

If the 5^{th} term of the A.P. is 31 and 25^{th} term is 140 more than the 5^{th} term, find the A.P.

**Solution:**

Let a be the first term and d be the common difference.

We know that, nth term (a_{n}) = a + (n – 1)d

According to question, a_{6} = 31

⟹ a + (5 – 1) = 31

⟹ a + 4d = 31

⟹ a = 31 – 4d …. (1)Also, a_{25 } = 140 + a_{5}

⟹ a + (25 – 1) = 140 + 31

⟹ a + 24d = 171 …. (3)

On substituting the values of (1) in (2), we get

31 – 4d + 24d = 171

⟹ 20d = 171 – 31

⟹ 20d = 140

⟹ d = 7

⟹ a = 31 – 4 × 7 [From (1)]

⟹ a = 3 Thus, the A.P. obtained is 3, 10, 17, 24, . .

**All Chapter RD Sharma Solutions For Class10 Maths**

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