Ex-9.4 Arithmetic Progressions (Part - 3), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 31. 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 32. 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,

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 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 = 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

2/5 d + 4d = 22

⇒ 2d + 20d = 22 X 5

⇒ 22d = 110

⇒ d = 5

Now,  a = 2/5 X 5                                                   [From (1)]

⇒ a = 2

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

Question 34. 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,

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,

a­₇₂  = 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, 72^nd term of the given A.P. is 4 times its 15^th term.

Question 35. 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,

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

Question 36. If the seventh 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,

a₇ = 1/9

⇒ a + (7 – 1)d = 1/9

⇒ a + 6d = 1/9                                                                        ….(1)

Also, a₉ = 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 X 1/63 = 1/9

⇒ a = 1/9  –  6/63

⇒ a = (7 – 6)/63

⇒ a = 1/63

Therefore, a₆₃  = a + (63 – 1)d

= 1/63  +  62/63

= 63/63

= 1

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

Question 37. 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,

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 = 3/2.d                                                                         ….(2)

Substituting the value of (2) in (1), we get 3/2.d + 6d = 15

⇒  3/2.d  + 6d = 15

⇒ 3d + 12d = 15 x 2

⇒ 15d = 30

⇒ d = 2

now, a = 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, . . .

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 = 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: 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.

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 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,

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, . .