Telangana SCERT Class 10 Maths Solution Chapter 6 Progressions Exercise 6.2

Telangana SCERT Solution Class X (10) Maths Chapter 6 Progressions Exercise 6.2

 

Exercise – 6.2

 

(Q1) Fill in the blanks in the following table, given that a is the first term, d the common difference and a_∩ the ∩^th term of the A.P.

SL No.	a	d	n	an
(i)	7	3	8	28
(ii)	-18	2	10	0
(iii)	46	-3	18	-5
(iv)	-18.9	2.5	10	3.6
(v)	3.5	0	105	3.5

(i) a = 7, d = 3, n = 8, a_n =?

=> Solution:

In an A.P,

Given: a = 7, d = 3, n = 8,

a_n = a + (n – 1) d

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a₈ = 7 + (8-1) 3

= 7 + 7 X (3)

a₈ = 7 + 21

= 28

(ii) Given : a = – 18, n = 10, a_n = 0, d =?

In an A.P.

An = a + (n – 1) d

0 = – 18 + (10 – 1) X d

0 = – 18 + 9d

0 = – 18 + 9d

18 = 9d

18/9 = d

d = 2

(iii) Given: a =?, d = -3, n = 18, a_n = – 5

In an A.P.

a_n = a + (n – 1) d

– 5 = a + (18 – 1) (-3)

– 5 = a + 17 (-3)

– 5 + 51 = a

46 = a

∴ a = 46

(iv) Given: a = 18.9, d = 2.5, n =? an = 36

In an A.P

an = a + (n – 1) d

3.6 = -18.9 + (n – 1) 2.5

3.6 + 18.9 = (n – 1) 2.5

22.5 = (n – 1) 2.5

22.5/2.5 = n – 1

9 = n – 1

9 + 1 = n

n = 10

(v) a = 3.5, d = 0, n = 105, a_n =?

=> Given: a = 3.5, d = 0, n = 105

a_n = a + (n – 1) d

= 3.5 + (105 – 1) 0

= 3.5 + (104) 0

= 3.5 + 0

a_n = 3.5

(Q2) Find the

(i) 30^th term of the A.P, 10, (1’1)

=> Solution:

The given A.P is

10, 7, 4, —–

A = 10, d = 7 – 10 = – 3

We have, an = a + (n – 1) d

a₃₀ = 10 + (30 – 1) (-3)

= 10 + 29 (-3)

a₃₀ = – 77

(ii) 11^th term of the A.P – 3, -1/2, 2

=> Solution:

The given A.P is,

– 3, -1/2, 2, —-

A = -3, d = -1/2 – (-3) = -1/2 + 3 = -1+6/2 = 5/2

We have, a_n = a + (n – 1) d

a₁₁ = – 3 + (11 – 1) (5/2)

a₁₁ = -3 + (10) (5/2)

= -3 + 5 (5)

= – 3 + 25

a₁₁ = 22

(Q3) Find the respective term for the following A.P.S

(i) a₁ = 2, a₃ = 26 find a₂

=> Solution:

Given: a₁ = a = 2 and a₃ = 26

a₃ = 26

a + 2d = 26 [∵ a₃ = a + 2d)

2 + 2d = 26

2d = 26 – 2

2d = 24

d = 24/2

d = 12

a₂ = a + d = 2+12 = 14

a₃ = a + 2d

= 18 + 2 (-5)

= 18 – 10

a₃ = 8

∴ a₁ = 18 and a₃ = 8

(iii) a₁ = 5, a₄ = 9 1/2 find a₂, a₃

=> Solution:

Given: a₁ = 5, a₄ = 9 1/2

a₁ = a = 5

a₄ = 9 1/2 = 19/2

a + 3d = 19

5 + 3d = 19/2

3d = 19/2 – 5

= 19-10/2

3d = 9/2

d = 9/3X2 = 9/6

d = 3/2

a₂ = a + d

= 5+3/2

= 10+3/2

a₂ = 13/2

a₃ = a + 2d

= 5 + 2 (3/2)

= 5 + 3

a₃ = 8

∴ a₂ = 13/2 and a₃ = 8

(iv) a₁ = – 4, a₆ = 6 find a₂, a₃, a₄, a₅

=> Solution:

Given: a₁ = – 4, a₆ = 6

a₁ = a = -4

a + 5d = 6   (a₆ = a + 5d)

– 4 + 5d = 6

5d = 6+4

5d = 10

d = 10/5

d = 2

a₂ = a + b

= – 4 + 2

a₂ = – 2

a₃ = a + 2b

= – 4 + 2 (2)

= – 4 + 4

a₃ = 0

a₄ = a + 3d

= – 4 + 3 (2)

= – 4 + 6

a₄ = 2

a₅ = a + 4d

= – 4 + 4 (2)

= – 4 + 8

a₅ = 4

∴ a₂ = -2, a₃ = 0, a₄ = 2 and a₅ = 4

a₃ = a + 2d

= 53 + 2 (-15)

= 53 – 30

a₃ = 2.3

a₄ = a + 3d

= 53 + 3 (-15)

a₄ = 53 – 45

a₄ = 8

a₅ = a + 4d

= 53 + 4(-15)

a₅ = 53 – 60

a₅ = – 7

∴ a₁ = 53, a₃ = 23, a₄ = 8 and a₅ = – 7

(Q1) Which term of the A.P. 3, 8, 13, 18

=> Solution:

Given A.P is 3, 8, 13, 18,

Here, a = 3, d = a₂ – a₁

= 8 – 3

= 5

Let a_n = 78

A_n = a + (n – 1) d

78 = 3 + (n – 1) 5

78 – 3 = (n – 1) 5

75 = (n – 1) 5

75/5 = n – 1

15 = n – 1

15 + 1 = n

16 = n

∴ n = 16

∴ 78 is the 16^th term.

(Q5) Find the number of terms in each of the following A.P.s

(i) 7, 13, 19 ——, 205

=> Solution: a = 7, d = a₂ – a₁, = 13 – 7 = 6

Let, a_n = 205

a_n = a + (n – 1) d

205 = 7 + (n – 1) 6

205 – 7 = (n – 1) 6

198 = (n – 1) 6

198/6 = n – 1

33 = n – 1

33+1 = n

34 = n

∴ n = 34

∴ There are 34 terms

(ii) 18, 15, 15 1/2, 13 —– -47

=> Solution:

a₁ = 18, d = a₂ – a₁ = 15-18/2 = -5/2

Let a_n = -47

a_n = a + (n – 1) d

– 47 = 18 + (n – 1) (-5/2)

-47 – 18 = (n – 1) (-5/2)

-65 = (n – 1) (-5/2)

-65 X (-2/5) = n – 1

13 X 2 = n – 1

26 = n – 1

26 + 1 = n

27 = n

∴ n = 27

∴ There are 27 terms

(Q6) Check whether, – 150 is a term of the A.P 11, 8, 5, 2

=> Solution: Given A.P 11, 8, 5, 2,

A = 11, d = 8 – 11 = – 3

Let a_n = – 150

a_n = a + (n – 1) d

– 150 = 11 + (n – 1) (-3)

– 150 – 11 = (n – 1) (-3)

– 161 = (n – 1) (-3)

– 161/-3 = n – 1

161/3 = n – 1

161/3 + 1 = n

161/3 + 3 = n

164/3 = n

∴ n = 164/3

∴ – 150 is not a term of given A.P.

Substituting d = – 2 in equⁿ (1)

a + 2d = 4

a + 2 (-2) = 4

a + (-4) = 4

a + 4 = 4

a = 4 + 4

a = 8

Let a_n = 0

a_n = a + (n – 1) d

0 = 8 + (n – 1) d

0 – 8 = 8 + (n – 1) (-2)

– 8 = (n – 1) (-2)

– 8 = (n – 1) (-2)

-8/-2 = n – 1

4 = n – 1

4 + 1 = n

∴ n = 5

∴ 5^th term of the A.P. is ‘0’

(Q9) The 17^th term of an A.P exceeds its 10^th term by 7. Find the common difference.

=> Solution:

Given: 17^th term of a_n A.P exceeds it is 10^th term by 7.

a₁₇ = a₁₀ + 7

=> a + 16d = a + 9d + 7

=> 16d – 9d = 7

=> 7d = 7

=> d = 7/7

D = 1

∴ Common difference = 1

(Q10) Two A.P.s have the same common difference. The difference between their 100^th terms is 100. What is the difference between their 100^th terms?

=> Solution:

Let the 100^th terms of the two A.P.s are

a₁ + 99d₁ and a₂ + 99d₂

Difference between them is 100

(a₂ + 99d₂) – (a₁ + 99d₁) = 100

a₂ + 99d₂ – a₁ – 99d₁ = 100

Since their common difference is same i.e. d₁ = d₂

a₂ + 99d₁ – a₁ – 99d₁ = 110

a₂ – a₁ = 100 —— (1)

Their 1000^th terms are,

a₁ + 99d₁ and a₂ + 99d₂

Difference between from

= (a₂ + 999d₂) – (a₁ + 999d₁)

= a₂ + 999d₂ – a₁ – 999d₁

= a₂ + 999d₁ – a₁ – 999d₁ (d₂ = d₁)

= a₂ – a₁

= 100 (from (10))

(Q11) How many three digit numbers are divisible by 72

=> Solution:

The three digit numbers = 100, 101, 102, 103, 104, —– 999

Since the common difference is same it forms an A.P.

Here, a = 105, d = 112 – 105 = 7

Let a_n = 994

a + (n – 1) d = 994

105 + (n – 1)d = 994

(n – 1) d = 994 – 105

(n – 1) 7 = 889

(n – 1) = 889/7

(n – 1) = 127

n = 127 + 1

n = 128

∴ There are 128 three digit numbers which are divisible by 7.

(Q12) How many multiples of 4 lie between 10 and 250?

=> Solution:

Multiple of ‘4’ between and 250 à 12, 16, 20, 24, 248

Since the common difference is same. It forms an A.P.

A = 12, d = a₂ – a₁ = 16 – 12 = 4

Let a_n = 248

A + (n – 1) d = 248

12 + (n – 1) 4 = 248

(n – 1) 4 = 248 – 12

(n – 1) 4 = 236

n – 1 = 236/4

n – 1 = 59

n = 59+1

n = 60

There are 60 multiples of 4 lie in between 10 and 250.

(Q13) For what value of n, are the n^th terms of two A.P.s 63, 65, 61

=> Solution:

The first A.P is 63, 65, 67 —–

a = 63, d = a₂ – a₁ = 65 – 63 = 2.

n^th term a_n = a + (n – 1) d

a_n = 63 + (n – 1) 2 ——- (1)

The second A.P is 3, 10, 17, ——

A = 3, d = a₂ – a₁ = 10 – 3 = 7

n^th term a_n = a + (n – 1) d

a_n = 3 + (n – 1) 7 ——– (2)

Since the n^th terms are equal from equations (1) and (2)

63 + (n – 1) = 3 + (n – 1) 7

63 – 3 = (n – n7 – (n – 1) 2

60 = 7n – 7 – 2n + 2

60 = 5n – 5

60 + 5 = 5n

65 = 5n

65/5 = n

13 = n

∴ n = 13

For n = 13, the n^th terms of the given two A.P.s are equal.

(Q14) Determine the A.P. whose third term is 16 and the 7^th term exceeds the 5^th term by 12.

=> Solution:

The 3^rd term is 16

a₃ = 16

a + 2d = 16 ——- (1)

7^th term exceeds the 5^th term by 12.

i.e. a₇ = a₆ + 12

a + 6d = a + 4d + 12

6d – 4d = 12

2d = 12

D = 12/6

D = 6

Substitute d = 6 in equⁿ (1), we get

a + 2 X 6 = 16

a + 11 = 16

a = 16 – 12

a = 4

a₂ = a + d = 4+6 = 10

a₃ = a+2d = 4 + 2(6) = 4 + 12 = 16

The A.P is 4, 10, 16, 22, ——

(Q16) The sum of the 4^th and 8^th terms of an A.P is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the A.P

=> Solution:

Sum of 4^th and 8^th terms is 24.

a₄ + a₈ = 24

a + 3d + a + 7d = 24

2a + 10d = 24

2 (a + 5d) = 24

a + 5d = 24/2

a + 5d = 12 ——- (1)

a₁ = a = -13

a₂ = a + d = -13 + 5 = -8

a₃ = a + 2d = -13 + 2(5) = -13 +10 = -3

The first three terms of the A.P are,

a₁ = -13, a₂ = -8, a₃ = -3

-13, -8 and -3

(Q15) Find the 20^th term from the end of the A.P 3, 8, 13, —— 253

=> Solution:

The given A.P. is

3, 8, 13, —– 253

a = 3, d = a₂ – a₁ = 8 – 3 = 5

Let a_n = 253

a + (n – 1) d = 253

3 + (n – 1) d = 253

3 + (n – 1) 5 = 253

(n – 1) 5 = 253 – 3

(n – 1) 5 = 250

n – 1 = 250/5

n – 1 = 50

n = 50 + 1

n = 51

There are 51 terms

20^th term from the end = (51 – 20) + 1

= 31 + 1

= 32^nd term from the first

a₃₂ = a + 31d

= 3 + 31(5)

= 3 + 155

a₃₂ = 158

(Q17) Subba Rao started this job in 1995 at a monthly salary of RS 5000 and received an increment of RS 200 each year. In which year did his salary reach RS 7000?

=> Solution:

Subba Rao in 1995 in RS 5000

Rearly increment is RS 200

List of his salaries are RS 5000, 5200, 5400, —–

Since the common difference is done it forms an A.P

a = 5000, d = 5200 – 5000

d = 200

Let a_n = 7000

a + (n – 1) d = 7000

5000 + (n – 1) 200 = 7000

(n – 1) 200 = 7000 – 5000

(n – 1) 200 = 2000

n – 1 = 2000/200

n – 1 = 10

n = 10 + 1

n = 11

After 11 years, his salary becomes RS 7000 i.e. 1995 + 11 = 2006

In the year 2006 his salary becomes RS 7000.

 

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