Ex-9.4 Arithmetic Progressions (Part - 2), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

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

(i) 1,21/11,31/11,41/11,....:x=141/11

(ii) 5(1/2),11,16(1/2),22,...:x=550

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

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

Solution:

(i) Given sequence is

1,21/11,31/11,41/11,....:x=141/11

first term (a) = 1

Common difference (d)  = 21/11 – 1

= (21 – 11)/11

= 10/11

n^th term a_n = a + (n - 1) X d

⇒  171/11=1 + (n – 1).10/11

⇒  171/11 – 1=(n – 1).10/11

⇒ (171 – 11)/11=(n – 1).10/11

⇒  160/11=(n – 1).1011

⇒  (n – 1)=160/11X11/10

⇒ n = 17

(ii) Given sequence is

5(1/2),11,16(1/2),22,...:x=550

first term  (a) = 5(1/2) = 11/2

Common Difference (d) = 11 – 11/2 = 11/2

n^th term a_n = a + (n - 1) X d

⇒  550 = 11/2 + (n – 1).11/2

⇒  550 = 11/2[1 + n – 1]

⇒ n = 550 X 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

n^th term a_n = 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

n^th term a_n = a + (n - 1) X d

we have a_n = 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 n^th term 1. Show that the sum of     the m^thterm 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 17. Find the A.P. whose third term is 16 and seventh term exceeds its fifth term by 12.

Solution: Given, a₃ = 16

⇒ 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 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 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

So, the obtained A.P. is

2, 7, 12, 17, . . .

Question 19. 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 + a₁₃

⇒ a + (n – 1)d = a + (13 – 1)d + 84

⇒ (n – 1) X 7 = 12 X 7 + 84

⇒ n – 1 = 24

⇒ n = 25

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

Question 20. 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 a₁, a₂ , a₃ , . . .  and  b₁, b₂ , b₃, . . .

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

⇒ a₁₀₀  – b₁₀₀ = 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

⇒ a₁₀₀₀  – b­₁₀₀₀

= 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 21. 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 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

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 23. 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) 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 41^st term?

Solution: Given sequence

8, 14, 20, 26, . . .

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

⇒ a_n  = a₄₁   + 72                                                                                 … (1)

For the given sequence,

first term (a) = 8,

Common Difference (d) = 14 – 8 = 6

from equation (1), we have

a_n  = 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, 53^rd term is 72 more than its 41^st term.

Question 25. 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 + 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, 49^th term of the A.P. 39 more than its 36^th term.

Question 26. 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 27. 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 X 2

= 126 – 18

= 108

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

Question 28. 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                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)

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

Question 30. 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

⇒ a₁₇  = 5 + 2a₈

⇒ 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

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 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