RD Sharma Solutions Class 10 Arithmetic Progressions Exercise 9.4

RD Sharma Class 10 Solutions Chapter 9 Ex 9.4 PDF Free Download

Exercise 9.4

Question 1. If 12th of an A.P is 82 and 18th term is 124. Then find out the 24th term.

Solution: Given: a12 = 82 and a18 = 124

General term of an AP : an = a + (n – 1)d

⇒ a12 = a + (12-1)d

⇒ 82 = a + 11d —(1)

⇒ 124 = a + (18-1)d

⇒ 124 = a + 17d —(2)

Subtracting (2) from (1)

⇒ ( a + 17d ) – ( a + 11d ) = 124 – 82

⇒ a +17d – a – 11d = 42

⇒ 6 d = 42

⇒ d = 7, common difference

Putting d = 7 in equation (1), we get

⇒ a + 11 X 7 = 82

⇒ a = 82 – 77

⇒ a = 5, first term

To find: 24th term

a24 = a + ( 24 -1 )d

[use value of a and d]

= 5 + 23 X 7

= 5 + 161

= 166

Question 2. In an A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Solution:

24th term is twice the 10th term (Given)

Which means, a24 = 2 X a10 . . . . . . (1)

Let, “a” be the first term and “d” be the common difference.

And, nth term is an = 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

To Prove: 72nd term is twice the 34th term

⇒ a72 = 2 X a34

⇒ 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)this twice the (m+n+1)th term.

Solution:

a(m+1) = 2 a(n+1) (From the question)

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)

To Prove: a(3m+1) = 2 a(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 proved.

Question 4. If the nth term of the A.P. 9,7,5, … is same as the nth term of the A.P. 15, 12 , 9, … find n.

Solution:

First sequence is 9,7,5, …

First term (a) = 9 and Common Difference(d) = 9 – 7 = -2

nth term = a + (n-1)d

⇒ an = 9 + (n-1)(-2)

= 9 – 2n + 2

= 11 – 2n

Second sequence is 15, 12, 9, …

here, First term (a) = 15

Common Difference (d) = 12 -15 = -3

nth term = a + (n-1)d

⇒ a’n = 15 + (n-1)(-3)

= 15 -3n +3

= 18 -3n

We are given in the question that the nth term of both the A.P.s are same,

So, a­n = a’n

⇒ 11 – 2n = 18 – 3n

⇒ n = 7

Question 5. Find the 13th term from the end in the following A.P.

(i). 4, 9, 14, … , 254.

Solution:

First term (a) = 4 and common difference (d) = 9 – 4 = 5

last term here (l) = 254

nth term from the end is : l – (n-1)d

we have to find 13th term from end then : l – 12d

= 254 – 12 X 5

= 254 – 60

= 194

(ii). 3, 5, 7, 9, …, 201.

Solution: we have,

First term (a) = 3 and common difference (d) = 5 – 3 = 2

last term here (l) = 201

nth term from the end is : l – (n-1)d

we have to find 13th term from end then : l – 12d

= 201 – 12 X 2

= 201 – 24

= 177

(iii). 1, 4, 7, 10, … , 88.

Solution:

First term (a) = 1 and common difference (c.d) = 4 -1 = 3

last term here (l) = 88

nth term from the end is : l – (n-1)d

we have to find 13th term from end then : l – 12d

= 88 – 12 X 3 = 88 – 36 = 52

Question 6. The 4th term of an A.P. is three times the first term and the 7th term exceeds the twice the third term by 1. Find the A.P.

Solution: As per the statement, 4th term of the A.P = thrice the first term

⇒ a4 = 3 first term

Using nth term formula, we have, a + ( 4 -1 )d = 3 a

Where, first term to be ‘a’ and the common difference be ‘d’

⇒ a + 3d = 3 a

⇒ a = \(\frac{3}{2}d\) —(1)

and also it is given that,

the 7th term exceeds the twice of the 3rd term by 1

⇒ a7 + 1 = 2 X a3

⇒ a + (7-1)d +1 = 2[ a + (3-1)d ]

⇒ a + 6d +1 = 2a + 4d

⇒ a = 2d + 1 —(2)

putting the value of a = \(\frac{3}{2}d\) from equation (1) in equation (2)

\(\frac{3}{2}d\) = 2d + 1

⇒ \(\frac{3}{2}d\) – 2d = 1

⇒ \(\frac{3d – 4d}{2} = 1\)

⇒ -d = 2

⇒ d = -2

put d = -2 in \(a = \frac{3}{2}d\)

⇒ a = \(\frac{3}{2}(-2)\)

⇒ 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 8th term and 13th term are 48 and 78 respectively.

Solution: Given: a8 = 48 and a13 = 78

nth term of an A.P. is: an = a + (n-1)d

so,

a8 = a + (8 – 1)d = a + 7d …………(1)

a13 = 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 X 6 = 48

⇒ a + 42 = 48

a = 4

Now, nth term will be: an = a + (n-1)d

= 4 + (n-1)6

= 4 + 6n – 6

an = 6n – 2

and the 3rd term will be

a3 = 6 X 3 – 2

an = 16

Question 8. How many three digit numbers are divisible with 3?

Solution: First and last 3 digit numbers which are divisible by 3 are 102 and 999 respectively.

So, here we have

First term (a) = 102

Common Difference (d) = 3

Last term (l) = 999

⇒ an = 999

⇒ a + (n – 1)d = 999

⇒ 102 + (n – 1)3 = 999

⇒ 102 + 3n – 3 = 999

⇒ 99 + 3n = 999

⇒ 3n = 900

⇒ n = 300

Therefore, there are 300 terms which are divisible by 3.

Question 9. An A.P. has 50 terms and the first term is 8 and the last term is 155. Find the 41st term from the A.P.

Solution:

First term (a) = 8

Number of terms (n) = 50

Last term (an) = 148

⇒ an = a + (n – 1)d

⇒ 155 = 8 + (50 – 1)d

⇒ 49 d = 147

⇒ d = 3

41st term will be: a + (41-1)d

⇒ 8 + 40 X 3 = 128

41st term = 128

Question 10. The sum of 4th and 8th term of an A.P. is 24 and the sum of the 6th and 10th 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 “d”

Given : 4th term + 8th term = 24

⇒ a4 + a8 = 24

⇒ ( a + (4 – 1)d ) + ( a + (8 -1)d ) = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24 …………(1)

Again, 6th term + 10th term = 34

⇒ a6 + a10 = 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 = \(\frac{5}{2}\)

Put d = \(\frac{5}{2}\) in equation (1)

⇒ 2a + 10 X \(\frac{5}{2}\) = 24

⇒ 2a + 25 = 24

⇒ 2a = -1

⇒ a = \(-\frac{1}{2}\)

Therefore, we have a = \(-\frac{1}{2}\) and d = \(\frac{5}{2}\)

Question 11. The first term of an A.P. is 7 and its 100th term is -488, Find the 50th term of the same A.P.

Solution: Given,

First term (a) = 7

100th term (a100) = -488

we know, an = a + (n – 1)d

⇒ (a100) = 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 50th term of the A.P.

Then, a50 = a + 49 d

= 7 + 49 X (-5)

= 7 – 245

= – 238

So, the 50th term of the A.P. is -238

Question 12. Find a40 – a30 of the following A.P.

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

Solution: 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 a40 – a30 = (a + 39d) – (a + 29d)

= a + 39 d – a -29 d

= 10 d

= 10 X 2

= 20

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

Solution: Given A.P. is 4, 9, 14, 19, . . .

Common difference ( d ) = 9 – 4 = 5

we have to find a40 – a30 = 10 d

= 10 X 5 = 50

Question 13. Write the expression am – an for the A.P. a, a + d, a + 2d, . . . Find the common differences for given expressions.

Solution: General Arithmetic Progression

a, a + d, a + 2d, . . .

am – an = ( a + (m – 1)d) – ( a + ( n – 1 )d )

⇒ a + md – d – a – nd +d

⇒ md – nd

am – an = (m – n) d ………….(1)

Let’s find the common difference of the A.P. for which

(i). 11th term is 5 and 13th term is 79

11th term ( a11 ) = 5

and 13th term (a­13) = 79

from equation (1),

taking m = 11 and n = 13

⇒ am – an = ( 13 -11 ) d

⇒ 79 – 5 = 2d

⇒ 74 = 2d

d = 37

(ii). a10 – a5 = 200

Solution: The difference between the 10th term and 5th term

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

⇒ a10 – a5 = ( 10 – 5 )d

⇒ 200 = 5 d

d = 40

(iii). 20th term is 10 more than the 18th term

Solution: Given,

a20 + 10 = a18

⇒ a20 – a18 = 10

from equation (1), we have

am – an = ( m – n ) d

⇒ a20 – a18 = ( 20 – 18 ) d

⇒ 10 = 2d

d = 5

Question 15. Find n if the given value of x is the n term. The given A.P.

(i) \(1,\frac{21}{11}, \frac{31}{11}, \frac{41}{11}, . . . . : x = \frac{141}{11}\)

(ii) \(5\frac{1}{2}, 11, 16\frac{1}{2}, 22, . . . : x = 550\)

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

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

Solution:

(i) Given sequence is

\(1, \frac{21}{11}, \frac{31}{11}, \frac{41}{11}, . . . . : x = \frac{141}{11}\)

first term (a) = 1

Common difference ( d) = \(\frac{21}{11} – 1\)

= \( \frac{21 – 11}{11}\)

= \(\frac{10}{11}\)

nth term an = a + (n-1) X d

⇒ \(\frac{171}{11} = 1 + (n – 1).\frac{10}{11}\)

⇒ \(\frac{171}{11} – 1 = (n – 1).\frac{10}{11}\)

⇒ \(\frac{171 – 11}{11} = (n – 1).\frac{10}{11}\)

⇒ \(\frac{160}{11} = (n – 1).\frac{10}{11}\)

⇒ \((n – 1) = \frac{160}{11} X \frac{11}{10}\)

⇒ n = 17

(ii) Given sequence is

\(5\frac{1}{2}, 11, 16\frac{1}{2}, 22, . . . : x = 550\)

first term (a) = \(5\frac{1}{2} = \frac{11}{2}\)

Common Difference ( d ) = \(11 – \frac{11}{2} = \frac{11}{2}\)

nth term an = a + (n-1) X d

⇒ \(550 = \frac{11}{2} + ( n – 1 ). \frac{11}{2}\)

⇒ \(550 = \frac{11}{2} [ 1 + n – 1 ]\)

⇒ n = 550 X \(\frac{2}{11}\)

⇒ 100

(iii) Given sequence is,

-1 , -3, -5, -7, . . . : x = -151

first term ( a ) = -1

Common Difference ( d ) = -3 – (-1)

= -3 + 1

= -2

nth term an = a + (n-1) X 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, . . . : x = 1000

First term ( a ) = 25

Common Difference ( d ) = 50 – 25 = 25

nth term an = a + (n-1) X d

we have an = 1000

⇒ 1000 = 25 + ( n – 1 ) 25

⇒ 975 = ( n – 1)25

⇒ n – 1 = 39

⇒ n = 40

Question 16. If an A.P. consists of n terms with the first term a and nth term 1. Show that the sum of the mth term from the beginning and the mth term from the end is (a + 1).

Solution: First term = a

Common Difference = d

Last term ( l ) = a + ( n – 1 ) d

Total no. of terms = n

mth term from the beginning am = a + ( n – 1 )d

mth term from the end = l + ( n – 1 )( -d )

⇒ a(n – m + 1) = l – (n – 1)d

⇒ am + a(n – m + 1) = a + (n – 1)d + ( l – ( n – 1)d )

= a + (n-1)d + l – (n-1)d

= a + l

Question 17. Find the A.P. whose third term is 16 and seventh term exceeds its fifth term by 12.

Solution: Given: a3 = 16 and a7 – 12 = a5

a3 = a + ( 3 – 1 )d = 16

⇒ a + 2d = 16 … (i)

and a7 – 12 = a5

⇒ 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 18. The 7th term of an A.P is 32 and its 13th term is 62. Find the A.P.

Solution: Given

a 7 = 32

⇒ a + ( 7 – 1 )d = 32

⇒ a + 6d = 32 … (i)

and a13 = 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 X 5 = 32

⇒ a = 32 – 30

⇒ a = 2

Answer: A.P. = 2, 7, 12, 17, . . .

Question 19. Which term of the A.P. 3, 10, 17, . . . will be 84 more than its 13th term?

Solution:

Given A.P. is 3, 10 , 17, . . .

First term ( a ) = 3

Common Difference (d) = 10 -3 = 7

Let nth term of the A.P. will be 84 more than its 13th term, then

an = 84 + a13

⇒ a + (n – 1)d = a + ( 13 – 1 )d + 84

⇒ ( n – 1 ) X 7 = 12 X 7 + 84

⇒ n – 1 = 24

⇒ n = 25

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

Question 20. Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?

Solution:

Let us consider,

First A.P. = a1, a2 , a3 , . . . and

Second A.P. = b1, b2 , b3, . . .

an = a1 + ( n – 1 )d and bn = a1 + ( n – 1 )d

Since, the difference between their 100th terms is 100.

⇒ a100 – b100 = 100

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

⇒ a + 99d – b – 99d = 100

⇒ a + b = 100 … (1)

Difference between 100th term is

⇒ a1000 – b­1000

= a + (1000 – 1)d – [ b + (1000 – 1)d ]

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

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

Question 21. For what value of n, the nth 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

nth term (an) = a + ( n – 1)d

= 63 + ( n – 1 ) 2

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

Common Difference ( d’ ) = 10 – 3 = 7

nth term ( a’n ) = a’ + ( n – 1 )d

= 3 + ( n – 1) 7

Let nth 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

The value of n is 13.

Question 22. 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

nth term ( an ) = a + ( n – 1)d

Here nth term a n = 248

⇒ 248 = a + ( n – 1 )d

⇒ 12 + (n – 1)4 = 248

⇒ (n – 1)4 = 236

⇒ n – 1 = 59

⇒ n = 59 + 1 = 60

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

Question 23. How many three digit numbers are divisible by 7?

Solution: All three digit numbers lies in the range: 100, . . . . . . , 999

Here 105 is the first 3 digit number which is divisible by 7.

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

The required sequence is

105, .. . . . . . . . , 994

Here, First term (a) = 105

Last term (l ) = 994

Common Difference (d) = 7

Let there are n numbers in the sequence then,

⇒ an = 994

⇒ a + ( n – 1 )d = 994

⇒ 105 + ( n – 1 )7 = 994

⇒ (n – 1) X 7 = 889

⇒ n – 1 = 127

⇒ n = 128

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

Question 24. Which term of the A.P. 8, 14, 20, 26, . . . will be 72 more than its 41st term?

Solution: Given sequence

8, 14, 20, 26, . . .

Let its n term be 72 more than its 41st term

⇒ an = a41 + 72 … (1)

For the given sequence,

first term (a) = 8,

Common Difference (d) = 14 – 8 = 6

from equation (1), we have

an = a41 + 72

⇒ a + ( n – 1 )d = a + ( 41 – 1 )d + 72

⇒ 8 + ( n – 1 )6 = 8 + 40 X 6 + 72

⇒ ( n – 1 )6 = 312

⇒ n – 1 = 52

⇒ n = 53

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

Question 25. Find the term of the Arithmetic Progression 9, 12, 15, 18, . . . which is 39 more than its 36th 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, an­ = 39 + a36

⇒ a + ( n – 1 )d = 39 + a + ( 36 – 1 )d

⇒ ( n -1 )3 = 39 + 35 X 3

⇒ ( n -1 )3 = ( 13 + 35 ) 3

⇒ n – 1 = 48

⇒ n = 49

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

Question 26. Find the 8th 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

nth term from end = l – ( n – 1 )d

8th term from end = 184 – ( 8 – 1 )3

= 184 – 7 X 3

= 184 – 21

= 183

Therefore, 8th term from the end is 183.

Question 27. Find the 10th 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

nth term from end is : l – ( n -1 )d

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

= 126 – 9 X 2

= 126 – 18

= 108

Therefore, 108 is the 10th term from the last in the gievn A.P.

Question 28. The sum of 4th and 8th term of an A.P. is 24 and the sum of 6th and 10th term is 44. Find the Arithmetic Progression.

Solution: Given: a4 + a8 = 24

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

⇒ 2a + 3d + 7d = 24

⇒ 2a + 10d = 24 …(1)

and a6 + a10 = 44

⇒ a + ( 6 – 1 )d + a + ( 10 – 1 )d = 44

⇒ 2a + 5d + 9 d = 44

⇒ 2a + 14d = 44 …(2)

Subtract equation (1) from equation (2), we get

2a + 14d – ( 2a + 10d ) = 44 -24

⇒ 4d = 20

⇒ d = 5

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

2a + 10X5 = 24

⇒ 2a = 24 – 50

⇒ 2a = -26

⇒ a = -13

The A.P is -13, – 7, -2, . . .

Question 29: 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 nth term is 120 more than 21st term

⇒ an = 120 + a21

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

⇒ ( n -1 )d = 120 + 20d

⇒ ( n – 1 )12 = 120 + 20 X 12

⇒ n – 1 = 10 + 20

⇒ n = 31

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

Question 30. The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43. Find the nth term.

Solution:

17th term of an A.P is 5 more than twice its 8th term

⇒ a17 = 5 + 2a8

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

⇒ a + 16d = 5 + 2a + 14d

⇒ a + 5 = 2d … (1)

and 11th term of the A>P. is 43

a11 = 43

⇒ a +( 11 – 1 )d = 43

⇒ a + 10d = 43

⇒ a + 5 X 2d = 43

from equation (1)

⇒ a + 5 X ( 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 nth term (an) = a + ( n – 1 )d

= 3 + ( n – 1 )4

= 3 + 4n – 4

= 4n – 1

Therefore, nth term is 4n – 1

Question 31. Find the number of all three digit natural numbers which are divisible by 9.

Solution: First 3-digit number that is divisible by 9 is 108.

The second number is 108 + 9 = 117.

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

Thus, the A.P. will be 108, 117, …. , 999.

Here, first term ( a ) = 108

last term (l) = 999

and the common difference ( d ) as 9

We know that, nth term ( an ) = 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 32. The 19th term of an A.P. is equal to three times its 6th term. if its 9th term is 19, find the A.P.

Solution: nth term formula for an A.P. is, an= a + (n – 1)d

Where “a” be the first term and “d” be the common difference.

According to the question,

a19 = 3a6

⇒ 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 x 2

⇒ 19d= 38

⇒d = 2

Now, a = 32×2 [From (1)]

a= 3

Therefore, The A.P. is : 3, 5, 7, 9, . . .

Question 33. The 9th term of an A.P. is equal to 6 times its second term. If its 5th 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 ( an ) = 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 = \(\frac{2}{5}\) …. (1)

Also, a5 = 22

⇒ a+(6 – 1)d= 22

⇒ a + 4d = 22 ….(2)

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

\(\frac{2}{5}\) d + 4d = 22

⇒ 2d + 20d = 22 X 5

⇒ 22d = 110

⇒ d = 5

Now, a = \(\frac{2}{5}\) X 5 [From (1)]

⇒ a = 2

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

Question 34. The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Solution: We know that,

nth term ( an ) = a + (n – 1)d

According to the question,

a24 = 2 a10

⇒ a + (24 – 1)d = 2(a + (10 – 1)d)

⇒ a + 23d = 2a + 18d

⇒23d – 18d = 2a – a

⇒ 5d = a

⇒ a = 5d …. (1)

Also,

72 = a + (72 – 1) d

= 5d + 71d [From (1)]

= 76d …. (2)

and

a15 = a + (15 – 1) d

= 5d + 14d [From (1)]

= 19d …. (3)

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

⇒ 76 d = 4 X 19 d

⇒ a72 = 4 X a15

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

Question 35. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

Solution: The number is divisible by both 2 and 5, 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.

first term = 110

and the common difference = 10.

We know that,

nth term = an= a + (n – 1)d

According to the question,

990 = 110 + ( n – 1 )10

⇒ 990 = 110 + 10n – 10

⇒ 10n = 990 – 100

⇒ n = 89

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

Question 36. If the seventh term of an A.P. is 1/9 and its 9th term is 1/7, find the 63rd term.

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

where “a” be the first term and “d” be the common difference.

According to the question,

a7 = \(\frac{1}{9}\)

⇒ a+(7 – 1)d = \(\frac{1}{9}\)

⇒ a+ 6d = \(\frac{1}{9}\) ….(1)

Also, a9 = \(\frac{1}{7}\)

⇒ a + (9 – 1)d = \(\frac{1}{7}\)

⇒ a + 8d = \(\frac{1}{7}\) ….(2)

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

⇒ 8d – 6d = \(\frac{1}{7}\) – \(\frac{1}{9}\)

⇒ 2d = \(\frac{9 – 7}{ 63 }\)

⇒ 2d = \(\frac{2}{63}\)

⇒ d= \(\frac{ 1 }{ 63 }\)

Put value of d = \(\frac{ 1 }{ 63 }\) in equation (1), we get

⇒ a + 6 X \(\frac{ 1 }{ 63 }\) = \(\frac{ 1 }{ 9}\)

⇒ a = \(\frac{ 1 }{ 9 }\) – \(\frac{ 6 }{ 63 }\)

⇒ a = \(\frac{ 7 – 6}{ 63 }\)

⇒ a = \(\frac{ 1 }{ 63 }\)

Therefore, a63 = a + ( 63 – 1 )d

= \(\frac{ 1 }{ 63 }\) + \(\frac{ 62 }{ 63 }\) = \(\frac{ 63 }{ 63 }\) = 1

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

Question 37. The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, Find the A.P.

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

where “a” be the first term and “d” be the common difference.

According to the question,

a5 + a9 = 30

⇒ a + (5 – 1)d + a + (9 – 1)d = 30

⇒ a + 4d + a + 8d = 30

⇒ 2a + 12d = 30

⇒ a + 6d = 15 …. (1)

Also, a25 = 3(a8)

⇒ a + (25 – 1)d = 3[a + (8 – 1)d]

⇒ a + 24d = 3a + 21d

⇒ 3a – a = 24d – 21d

⇒ 2a = 3d

⇒ a = \(\frac{ 3 }{ 2 } d\) ….(2)

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

⇒ \(\frac{ 3 }{ 2 } d\) + 6d = 15

⇒ 3d + 12d = 15 x 2

⇒ 15d = 30

⇒ d = 2

now, a = \(\frac{ 3 }{ 2 } d\) X 2 [From (1)]

⇒ a = 3

Therefore, the A.P. is 3, 5, 7, 9, . . .

Question 38. Find whether 0 (zero) is a term of the A.P. 40, 37, 34, 31, . . . or not.

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

where “a” be the first term and “d” be the common difference.

Given: a = 40, d = -3

Suppose there is a term whose value is zero, an= 0

According to the question,

⇒ 0 = 40 + (n – 1)(-3)

⇒ 0 = 40 – 3n + 3

⇒ 3n = 43

⇒ n = \(\frac{ 43 }{ 3 }\) …. (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 39. Find the middle term of the A.P. 213, 205, 197, . . . 37.

Solution:

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

where “a” be the first term and “d” be the common difference.

Given: a = 213,

d = -8

and an = 37

According to the question,

⇒ 37 = 213 + (n – 1)(-8)

⇒ 37 =213 – 8n+ 8

⇒ 8n = 221 – 37

⇒ an = 184

⇒ n=23 ….(1)

Therefore, the total number of terms are 23.

Let us say, there is an odd number of terms.

So, Middle term will be the 12th term.

a12 =213 + (12 – 1)(-8)

a12 = 213 – 88

= 125

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

Question 40. If the 5th term of the A.P. is 31 and 25th term is 140 more than the 5th term, 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 question,

a6 = 31

⇒ a + ( 5 – 1 ) = 31

⇒ a + 4d = 31

⇒ a = 31 – 4d . . . .(1)Also, a25 = 140 + a5

⇒ 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 X 7 [From (1)]

⇒ a = 3

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

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