TN samacheer kalvi 10th Maths | Chapter 2 - Numbers and Sequences | Exercise 2.5 solutions

TN samacheer kalvi 10^th Maths: Chapter 2 Numbers and Sequences – Exercise 2.5 solutions

1.     Check whether the following sequences are in A.P.

i)                 a – 3,   a – 5,  a – 7,…….

      

t₂ – t₁ = a – 5 – (a – 3)

 = a – 5 – a + 3

 = -2

 

                t₃ – t₂ = a – 7 – (a – 5)

                         = a – 7 – a + 5

                         = -2

 t₂ – t₁ = t₃ – t₂

 

∴Common difference is same between the consecutive terms. So the sequence is in A.P

 

ii)            1/2,  1/3, 1/4, 1/5,…..

t₂ – t₁ = 1/3  -  1/2

      = (2-3)/6

      = -1/6

 

t₃ – t₂ = 1/4 - 1/3

      = (3-4)/12

      = -1/12

 

t₂ – t₁ ≠ t₃ – t₂

 

∴Common difference is not same between the consecutive terms. So the sequence is not in A.P

 

iii)         9,  13,  17,  21,  25,...

t₂ – t₁ = 13 – 9 = 4

t₃ – t₂ = 17 – 13 = 4

t₄ – t₃ = 21 – 17 = 4

t₅ – t₄ = 25 – 21 = 4

∴Common differences are same between the consecutive terms. So the sequence is in A.P

 

iv)          -1/3,   0,  1/3,   2/3

   

t₂ – t₁ = 0 –(-1/3)

 = 0+(1/3)

      = 1/3

 

t₃ – t₂ = (1/3) – 0

      = 1/3

 

t₄ – t₃ = (2/3) – (1/3)

      = (2-1) / 3

      = 1/3

 

∴Common differences are same between the consecutive terms. So the sequence is in A.P

 

v)             1,  -1,   1, -1,  1, -1,...

 

t₂ – t₁    = -1 – 1 = -2

t₃ – t₂    = 1 – (-1) 

             = 1 + 1 = 2

t₄ – t₃    = -1-(1) 

             = – 1 – 1 = – 2

t₅ – t₄   = 1 – (-1)

              = 1 + 1 = 2

 

∴Common differences are not same between the consecutive terms. So the sequence is not in A.P

YouTube video link: Ex2.5 question 1

 

2.     First term a and common difference d are given below. Find the corresponding A.P. ?

i)               a = 5 , d = 6

The general form of the A.P is a, a + d, a + 2d, a + 3d….

= 5,  5+6,  5+(2x6),   5+(3x6),…..

= 5, 11,  5+12,   5+18,…..

 

∴The A.P. 5, 11, 17, 23 ….

 

ii)            a = 7 , d =−5

The general form of the A.P is a, a + d, a + 2d, a + 3d….

= 7,  7+(-5),  7+[2x(-5)],   7+[3x(-5)], …..

= 7, 2, -3, -8, …..

 

∴The A.P. 7, 2, -3, -8,…..

 

iii)         a=3/4,   d=1/2

The general form of the A.P is a, a + d, a + 2d, a + 3d….

                

         = 3/4,  (3/4)+(1/2),    (3/4)+ [2x(1/2)],   (3/4)+ [3x(1/2)], ….

                

         = 3/4,  (3/4)+(2/4),    (3/4)+ (4/4),   (3/4)+(6/4),…..

 

         =3/4, (3+2)/4,  (3+4)/4,  (3+6)/4,…..

                

∴The A.P. 3/4, 5/4,  7/4, 9/4,....

     YouTube video link: Ex2.5 question 2

3.     Find the first term and common difference of the Arithmetic Progressions whose n^th terms are given below

i)               t_n = -3 + 2n  

  

When, n=1,

t₁ = -3+ 2(1)

   = -3+2

   = -1

   When, n=2,

t₂ = -3+ 2(2)

   = -3+4

   = 1

First term (a)=t1= -1

Common difference (d) = t₂ – t₁ 

                                     = 1-(-1)

                                     = 1+1

                                     = 2

 

ii)            t_n = 4 - 7n  

  

When, n=1,

t₁ = 4- 7(1)

   = 4-7

   = -3

   When, n=2,

t₂ = 4-7(2)

   = 4-14

   = -10

 

First term (a)=t1= -3

Common difference (d) = t₂ – t₁ 

                                     = -10-(-3)

                                     = -10+3

                                     = -7

YouTube video link: Ex2.5 question 3

4.     Find the 19th term of an A.P. -11,-15,-19,...

Solutions:

First term (a)=-11

Common difference (d) = t₂ – t₁ 

                                              =-15-(-11)

                                              =-15+11

                                              =-4

t₁₉ =? _, n=19

                       

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

                

                 t₁₉     =  -11 + [(19-1)x(-4)]

                          =  -11+[18x(-4)]

                          =  -11+(-72)

                 t₁₉     =  -83

    YouTube video link: Ex2.5 question 4

 

5.     Which term of an A.P. 16, 11, 6, 1,... is -54 ?

Solutions:

First term (a)=16

Common difference (d) = t₂ – t₁ 

                                              =11-16

                                              =-5

         Last term (l)=-54

 

                 No.of terms n =  [(l-a)÷d] + 1

        

         n       =  [ (-54-16) ÷ (-5) ] + 1

                 =  [ (-70) ÷ (-5) ] + 1

                  =  14 + 1

                 =  15

            ∴ 15th term is -54.

YouTube video link: Ex2.5 question 5

 

6.     Find the middle term(s) of an A.P. 9, 15, 21, 27,…,183.

Solutions:

First term (a)=9

Common difference (d)  = t₂ – t₁ 

                                                     =15-9

                                                = 6

         Last term (l)=183

 

                 No.of terms n =  [(l-a)÷d] + 1

 

         n      =  [ (183-9) ÷ 6 ] + 1

                 =  [ 174 ÷ 6 ] + 1

                 =  29 + 1

                 = 30

∴Middle terms are 15^th & 16^th terms

 

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

 

                 t₁₅     =  9 + [(15-1)x6]

                          =  9 + [14x6]

                          =  9 + 84

                          =  93

                  ∴Middle terms are t₁₅ = 93 or t₁₆ = 99

        YouTube video link: Ex2.5 question 6          

7.     If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.

Solutions:

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

Nine times ninth term = Fifteen times fifteenth term

9t₉ = 15t₁₅

9(a + (9-1)d) = 15(a + (15-1)d)

9(a + 8d) = 15(a + 14d)

9a + 72d = 15a + 210d

15a + 210d – 9a – 72d = 0

⇒ 6a + 138 d = 0

⇒ 6(a + 23 d) = 0

⇒ 6(a + (24 – 1)d) = 0

⇒ 6t₂₄ = 0. Hence it is proved.

   YouTube video link: Ex2.5 question 7

8.     If 3 + k, 18 – k, 5k + 1 are in A.P. then find k?

Solutions:

3 + k, 18 – k, 5k + 1 are in AP

∴ t₂ – t₁ = t₃ – t₂ (common difference is same)

18 – k – (3 + k) = 5k + 1 – (18 – k)

18 – k – 3 – k = 5k + 1 – 18 + k

15 – 2k = 6k – 17

32 = 8k

k = 32÷8 = 4

The value of k = 4

   YouTube video link: Ex2.5 question 8

9.     Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.

Solutions:

                  ∴ t₂ – t₁ = t₃ – t₂ = t₄ – t₃ =  t₅ – t₄

          t₂ – t₁ = 10 – x …….. (1)

                   t₃ – t₂ = y – 10……….. (2)

                   t₄ – t₃ = 24 – y ............(3)

                   t₅ – t₄ = z – 24 …….. (4)

From (2) and (3)

⇒ y – 10 = 24 – y

2y = 24 + 10 = 34

y = 34÷2 = 17

From (1) and (2)

⇒ 10 – x = y – 10

10 – x = 17 – 10 = 7

-x = 7 – 10

-x = -3 ⇒ x = 3

From (3) and (4)

24 – y = z – 24

24 – 17 = z – 24

7 = z – 24

z = 7 + 24 = 31

∴ Solutions x = 3, y = 17, z = 31

YouTube video link: Ex2.5 question 9

10. In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?

Solutions:

Number of seats in the first row

    (a)  = 20

    ∴ t₁= 20

Number of seats in the second row

    (t₂) = 20 + 2

          = 22

Number of seats in the third row

    (t₃) = 22 + 2

          = 24

Here a = 20 ; d = 2

Number of rows

     (n) = 30

n^th term (t_n) = a + (n-1)d

t₃₀     = 20 + (30-1)x(2)

         = 20 + 29x2

         = 20+58

t₃₀     = 78

∴ Number of seats in the last row is 78

YouTube video link: Ex2.5 question 10

 

11. The sum of three consecutive terms that are in A.P. is 27 and their product is 288.Find the three terms.

Solutions:

Let the three consecutive terms be a – d, a, a + d

Their sum = a – d + a + a + d = 27

                                           3a = 27

                                             a = 27÷3 

                                            a = 9

Their product = (a – d)(a)(a + d) = 288

⇒ a(a-d)(a+d) = 288

⇒ 9(a² – d²) = 288

⇒ 9(9² – d²) = 288

⇒ 9(81 – d²) = 288

⇒ 81 – d² = 288 ÷ 9

⇒ 81-d² = 32

⇒-d² = 32 – 81

⇒ d² = 49

⇒ d = ± 7

∴ The three terms are if a = 9, d = 7

⇒ a – d, a, a + d

= 9 – 7, 9, 9 + 7

∴ A.P. = 2, 9, 16

⇒if a = 9, d = -7

= 9 – (-7), 9, 9 + (-7)

=9+7, 9, 9-7

∴ A.P. = 16, 9, 2

YouTube video link: Ex2.5 question 11

12. The ratio of 6^th and 8^th term of an A.P. is 7:9. Find the ratio of 9^th term to 13^th term.

Solutions:

n^th term (t_n) = a + (n-1)d

 

Given : t₆ : t₈ = 7 : 9

[a + (6-1)d] : [a + (8-1)d] = 7 : 9

(a + 5d) : (a + 7d) = 7 : 9

9 (a + 5 d) = 7 (a + 7d)

9a + 45 d = 7a + 49d

9a – 7a = 49d – 45d

2a = 4d

a = 2d

To find t₉ : t₁₃

t₉ : t₁₃  = [a + (9-1)d] : [a + (13-1)d] 

= (a + 8d) : (a + 12d)

Sub. a=2d in above statement,

= (2d + 8d) : (2d + 12d)

= 10d : 14d

= 5 : 7

∴ t₉ : t₁₃ = 5 : 7

YouTube video link: Ex2.5 question 12

13. In a winter season let us take the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days.

Solutions:

Let the five days temperature be

(a – 2d), (a – d), a, (a + d) and (a + 2d)

Sum of first three days temperature = 0

a – 2d + a – d + a = 0

3a – 3d = 0

a – d = 0 …..(eq1)

Sum of the last three days temperature = 18°C

a + a + d + a + 2d = 18

3a + 3d = 18

(÷ by 3) ⇒ a + d = 6 ……(eq2)

By adding (eq1) and (eq2)

a   -     d      =    0

a   +    d      =    6

(+)__________________­­

2a   =   6

         a   =   6 ÷ 2

         a   =   3

Substitute to value of a = 3 in (eq2)

3 + d =  6

                             d    =  6-3

                             d    =  3

The temperature in 5 days are

(3 – 6), (3 – 3), 3, (3 + 3) and (3 + 6)

∴  -3°C, 0°C, 3°C, 6°C, 9°C

 YouTube video link: Ex2.5 question 13

14. Priya earned ₹15,000 in the first month. Thereafter her salary increases by ₹1500 per year. Her expenses are ₹13,000 during the first year and the expenses increases by ₹900 per year. How long will it take her to save ₹20,000 per month.

Tabulate the given table

Duration of the year	Monthly salary	Monthly expenses	Monthly savings
I year	15,000	13,000	2000
II year	16,500	13,900	2600
III year	18,000	14,800	3200

 

                Monthly savings form an A.P. = 2000,  2600,  3200 …..

                a = 2000

                d = 2600 – 2000 = 600

                last term (l) = 20,000

No.of terms n =  [(l-a)÷d] + 1

n = [(20,000-2000) ÷ 600] + 1

n = [18,000 ÷ 600] + 1

n = 30 + 1

                  n = 31

∴ He will take 31 years to save ₹ 20,000 per month

YouTube video link: Ex2.5 question 14