NCERT Grade 11-Sequences and Series-Answers

NCERT Solutions for Class 11 Maths

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1.    Given: an = n(n + 2)

Putting n = 1, 2, 3, 4 and 5, we get

an = n(n + 2) = 1 × 3 = 3

an = n(n + 2) = 2 × 4 = 8

an = n(n + 2) = 3 × 5 = 15

an = n(n + 2) = 4 × 6 = 24

an = n(n + 2) = 5 × 7 = 35

Therefore, the first five terms are 3, 8, 15, 24 and 35.

2.    Given: 

Putting n = 1, 2, 3, 4 and 5, we get,

Therefore, the first five terms are

3.    Given: an = 2

Putting n = 1, 2, 3, 4 and 5, we get,

a1 = 21 = 2

a2 = 22 = 4

a3 = 23 = 8

a4 = 24 = 16

a5 = 25 = 32

Therefore, the first five terms are 2, 4, 8, 16 and 32.

4.    Given: 

5.    Given: a = (-1) .5

Putting n = 1, 2, 3, 4 and 5, we get,

a1 = (-1)1-1 .51+1 = (-1)0 .52 = 1 × 25 = 25

a2 = (-1)2-1 52+1 = (-1)1 .53 = -1 × 125 = -125

a3 = (-1)3-1 .53+1 = (-1)2 .54 = 1 × 625 = 625

a4 = (-1)4-1 .54+1 = (-1)3 .55 = 1 × 3125 = -3125

a5 = (-1)5-1 .55+1 = (-1)4 .56 = 1 × 15625 = 15625

Therefore, the first five terms are 25, -125, 625, -3125 and 15625.

6.    Given: 

7.    Given: an = 4n – 3

∴ a17, = 4 × 17 – 3 = 68 – 3 = 65

a24 = 4 × 24 – 3 = 96 – 3 = 93

Therefore, 17th and 24th terms are 65 and 93 respectively.

8.   Given: 

Therefore, 7th term is

9.    Given: an = (-1)n-1 n3

∴ a9= (-1)9-1 × (9)3 = (-1)8 × 729 = 729

Therefore, 9th term is 729.

10.   Given: 

Therefore, 20th term is 

11.   an = 3, an = 3an-1 + 2 for all n > 1

Putting n = 2, 3, 4 and 5, we get,

a2 = 3a2-1 + 2 = 3a1 + 2 = 3 × 3 + 2 = 9 + 2 = 11

a3 = 3a3-1 + 2 = 3a2 + 2 = 3 × 11 + 2 = 33 + 2 = 35

a4 = 3a4-1 + 2 = 3a3 + 2 = 3 × 35 + 2 = 105 + 2 = 107

a5 = 3a5-1 + 2 = 3a4 + 2 = 3 × 107 + 2 = 321 + 2 = 323

Hence the first five terms are 3, 11, 35, 107, 323.

Therefore, corresponding series is 3 + 11 + 35 + 107 + 323 + …..

12.   Given: 

Putting n = 2, 3, 4 and 5, we get

13.   Given: a1 = a2 = 2, an = an-1 – 1, n > 2

Putting n = 3, 4 and 5 we get,

a3 = a3-1 – 1 = a2 – 1 = 2 – 1 = 1

a4 = a4-1 – 1 = a3 – 1 = 1 – 1 = 0

a5 = a5-1 – 1 = a4 – 1 = 0 – 1 = -1

Hence the first five terms are 2, 2, 1, 0, -1.

Therefore, corresponding series is 2 + 2 + 1 + 0 + (-1) + ……

14.   Given: a1 = a2 and an-1 + an-1, n > 2

Putting n = 3, 4, 5 and 6, we have

a3 = a3-1 + a3-2 = a2 + a1 = 1 + 1 = 2

a4 = a4-1 + a4-2 = a3 + a2 = 2 + 1 = 3

a5 = a5-1 + a5-2 = a4 + a3 = 3 + 2 = 5

a6 = a6-1 + a6-2 = a5 + a4 = 5 + 3 = 8

15.   Odd integers from 1 to 2001 are 1, 3, 5, 7, ….., 2001.

Here, a = 1, d = 3 – 1 = 2 and an = 2001

16.   According to question, series is 105, 110, 115, 120, ….., 995

Here a = 105, d = 110 – 105 = 5 and an = 995

17.   According to question, a = 2 and 

18.   Here, 

19.   Let a be the first term and d be the common difference of given A.P.

And 

Subtracting eq. (ii) from eq.(i), we get

Putting value of d in eq.(i), we get,

20.  Here a = 25, d = 22 – 25 = -3 and Sn = 116

Now, an = a + (n – 1)d

⇒ a8 = 25 + (8 – 1) × (-3)

⇒ a8 = 25 – 21 = 4.

21.  Given: ak = 5k + 1

Putting k = 1 and k = n, we get

a = 5 × 1 + 1 = 6 and an = 5n + 1

22.  Given: Sn = pn + qn2

Put n = 1 we get, S1 = p + q ⇒ a = p + q   …..(i)

Now Sn = pn + qn2

23.  Let a1, a2 and d1, d2 be the first terms and common differences of two A.P’s respectively.

Therefore, the ratio of 18th terms of two A.P.’s is 179 : 321.

24.  Let a be the first term and d be the common difference of given A.P.

According to question, Sp = Sq

25.  Let A be the first term and d be the common difference of given A.P.

Putting the values of  from eq.(i), (ii) and (iii), we get

⇒ L.H.S. = R.H.S. Proved.

26.  Let a be the first term and d be the common difference of given A.P.

27.  Given: Sn = 3n2 + 5n and am = 164

Put n = 1 and n = 2 in Sn we get, S1 = 3 + 5 = 8 ⇒ a = 8     …..(i)

and S = 3(4) + 5(2) = 12 + 10 = 22

We have a = S – S

∴ an = S2 – S1

⇒ a2 = 22 – 8 = 14

Now d = a2 – a1 = 14 – 8 = 6

∴ am = 164 ⇒ a + (m – 1)d = 164

⇒ 8 + (m – 1)6 = 164

⇒ 8 + 6m – 6 = 164

⇒ 6m = 162

⇒ m = 162/6 = 27

28.  Let A1, A2, A3, A4 and A5 be five numbers between 8 and 26 such that

8, A1, A2, A3, A4, A5, 26 are in A.P.

Here, a = 8 and a7 = 26 and let d be the common difference.

∴ a7 = a + (7 – 1)d = 26

⇒ 8 + 6d = 26

⇒ 6d = 18

⇒ d = 3

Now, A1 = a + d = 8 + 3 = 11

A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20

A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Hence the five numbers are 11, 14, 17, 20 and 23.

29.  Since, A.M. between a and b is 

30.  Let A1, A2, A3, A4, …….., Am be numbers between 1 and 31.

Here, a = 1 and let the common difference be d.

31.  Amount of 1st instalment = Rs. 100 and Amount of 2nd instalment = Rs. 105

The monthly instalments 100,105,110, ………….. form an A.P

∴ a = 100, d = 105 – 100 = 5 and n = 30

Now a = a + (n – 1)d

⇒ a30 = 100 + (30 – 1) × 5

⇒ a30 = 100 + 29 × 5

⇒ a30 = 100 + 145 = Rs. 245

Therefore, the amount of 30th instalments is Rs. 245.

32.   Let the number of sides of polygon be n. The interior angles of the polygon form an A.P.

Here, a = 120° and d = 5°

Since Sum of interior angles of a polygon with sides is (n – 2) × 180°

But n = 16 nit possible because a16 = a + 15d = 120 + 15 × 5 = 195° > 180°

Therefore, number of sides of the polygon are 9.

33.   

34.  Let a be the first term of given G.P. Here r = 2 and a8 = 192

35.  Let a be the first term and r be the common ratio of given G.P.

∴ a5 = p ⇒ ar4 = p         ..(i)

a8 = q ⇒ ar7 = q         ..(ii)

a11 = s ⇒ ar10 = s         ..(ii)

Squaring both sides of eq.(ii), we get a2 = (ar7)2

⇒ q2 = a2r14

⇒ q2 = (ar4) (ar10)

⇒ q2 = ps [from eq.(i) and (iii)]

36.  Let a be the first term and r be the common ratio of given G.P.

Here a = -3 and a4 = (a2)2

Now, a4 = (a2)2

⇒ ar3 = (ar)2

⇒ ar3 = a2r2

⇒ r = a

⇒ r = -3 [∵ a = -3]

∴ a7 = ar7-1 = (-3) × (-3)6

= -3 × 729 = -2187.

37.   (a) Here 

Therefore, 13th term of the given G.P. is 128.

Therefore, 12th term of the given G.P. is 729.

38.  Given:  are in G.P.

Therefore for x = ±1 the given numbers are in G.P.

39.   

40.  

41.  

42.  

43.  

Putting the value of Sn in eq.(i), we get 

44.  Let a/r, a, r be first three terms of the given G.P.

45.   

Therefore, the sum of 4 terms of the given G.P. is 120.

46.  Let a be the first term and r be the common ratio of given G.P.

∴ a + ar + ar2 = 16

⇒ a(1 + r + r2) = 16    …(i)

And ar3 + ar4 + ar5 = 128

⇒ ar3 (1 + r + r2) = 128    …(ii)

Putting the value from eq.(i) into eq.(ii) we get,

16r3 = 128

⇒ r3 = 8

⇒ r = 2

Putting value of r in eq(i), we get a (1 + 2 + 22) = 16

47.  Given: a = 729 a7 = 64

48.  Let a be the first term and r be the common ratio of given G.P.

Given: a + ar = -4

⇒ a(1 + r) = -4      …(i)

And a5 = 4a3

⇒ ar4 = 4ar2

⇒ r2 = 4

⇒ r = ±2

Putting r = 2 in eq.(i), we get a(1 + 2) = -4

Therefore, required G.P. is 

Putting r = -2 in eq.(i) we get a(1 – 2) = -4

⇒ a = 4

Therefore, required G.P. is 4, -8, 16, -32,…..

49.  Let a be the first term and r be the common ratio of given G.P.

∴ a4 = x

⇒ ar3 = x         …(i)

a10 = y

⇒ ar9 = y        …(ii)

a16 = z

⇒ ar15 = z       …(iii)

From eq.(ii), ar9 = y

⇒ (ar8)2 = y2

⇒ y2 = (ar3) (ar15)

⇒ y2 – xz [From eq.(i) and (iii)]

∴ x, y, z are in G.P.

50.  Here Sn = 8 + 88 + 888 + 8888 + …. up to n terms

51.  Multiplying the corresponding terms of the given sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

52.  Multiplying the corresponding terms of the given sequences, we have

Since the ration of the two succeeding terms are same, the resulting sequence is also in G.P. and common ratio 

53.  Let the four numbers in G.P. be a, ar, ar2, ar3

∴ ar2 = a + 9 and ar = ar3 + 18

Now, ar2 – a = 9

⇒ a(r2 – 1) = 9        …(i)

And ar – ar3 = 18

⇒ ar (1 – r2) = 18

⇒ -ar (r2 – 1) = 18    …(ii)

Dividing eq.(ii) by eq.(i), we have

Putting value of r in eq.(i), we get

a(4 – 1) = 9

⇒ a = 3

∴ ar = 3 × (2) = -6

ar2 = 3 × (-2)2 = 12

ar3 = 3 × (-2)3 = -24

Therefore,  the required numbers are 3, -6, 12, -24.

54.  Let A be the first term and R be the common ratio of given G.P.

55.  Let r be the common ration of the given G.P.

Here, first term of G.P. is a

and an = b

⇒ arn-1 = b       …(i)

Given: P = a, ar, ar2, ar3, ……..arn-1

⇒ P2 = a2nrn(n-1) = [aarn-1]n [Squaring both sides]

⇒ P2 = (ab)n [From eq.(i)]

Hence proved.

56.  Let a be the first term and r be the common ratio of given G.P.

57.  Given a, b, c, d are in G.P.

Let r be the common ratio of given G.P.

Then b = ar, c = ar2 and d = ar3

Now, L.H.S. = (a2 + b2 + c2) (b2 + c2 + d2)

= (a2 + a2r2 + a2r4) (a2r2 + a2r4 + a2r6)

= a2 (1 + r2 + r4) a2r2 (1 + r2 + r4) = a4r2 (1 + r2 + r4)2

R.H.S. = (ab + bc + cd)2

= (a.ar + ar.ar2 + ar2.ar3)2

= (a2r + a2r3 + a2r5)2

= (a2r)2 (1 + r2 + r4)2 = a4r2 (1 + r2 + r4)2

Therefore, L.H.S = R.H.S.

58.  Let G1 and G2 be two numbers between 3 and 81 such that 3, G1, G2, 81 are in G.P.

Let r be the common ration

Here a = 3 and a = 81

⇒ ar3 = 81

⇒ 3 × r3 = 81

⇒ r3 = 27

⇒ r = 3

∴ G1 = ar = 3 × 3 = 9

And G2 = ar2 = 3 × (3)2 = 27.

Therefore, the required numbers are 9 and 27.

59.  Since, G.M. between two numbers a and b is √ab.

60.  Let the numbers be a and b

Given: 

Applying componendo and dividendo, we get

Again applying componendo and dividendo, we get

Therefore, the numbers are in the ratio (3 + 2√2) : (3 – 2√2).

61.  Let the two positive numbers be a and b

62.  Bacteria present in the culture originally = 30

Since the bacteria doubles itself after each hour, then the sequence of bacteria after each hour is a G.P.

Here a = 30 and r = 2

∴ Bacteria at the end of 2nd hour = 30 × 23-1 = 30 × 22 = 120

And Bacteria at the end of 4th hour = 30 × 25-1 = 30 × 24 = 480

And Bacteria at the end of nth hour = an+1 = 30 (2(n+1)-1) = 30(2n)

63.  Original amount = Rs. 500, Rate of interest = 10% compounded annually

∴ Interest of one year   Rs. 50

And amount after one year = 500 + 50 = Rs. 550

Here a = 500 and 

Therefore, amount after 10 years = Amount in the 11th year = 500 × (1.1)11-1 = Rs. 500 (1.1)10

64.  Let a and b the roots of required quadratic equation.

Then A.M. 

⇒ a + b = 16

And G.M = √ab = 5

⇒ ab = 25

Now, Quadratic equation x2 – (Sum of roots)x + (Product of roots) = 0

⇒ x2 – (a + b)x + ab = 0

⇒ x2 – 16x + 25 = 0

Therefore, required equation is x2 – 16x + 25 = 0.

65.  Given: 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +…to n terms

∴ an = [nth term of 1, 2, 3, …….] [nth term of 2, 3, 4, 5, …….]

66.  Given: 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +…to n terms

∴ an = [nth term of 1, 2, 3, …….] [nth term of 2, 3, 4, 5, …….] [nth term of 3, 4, 5, …….]

67.  3 × 12 + 5 × 22 + 7 × 32 + ……to n terms

∴ an = [nth term of 3, 5, 7, …….] [nth term of 1, 2, 3, 4, 5, …….]2

= (2n + 1) (n)2 = 2n3 + n2

68.  

Then 1 = A(n + 1) + Bn

Put n = 0 then A = 1

Put n = -1 then B = -1

69.  Given: 52 + 62 + 72 + …… + 202

70.  3 × 8 + 6 × 11 + 9 × 14 + …..to n terms

∴ an = [nth term of 3, 6, 9, …….] [nth term of 8, 11, 14, ]

71.  12 + (12 + 22) + (12 + 22 + 32) +…..to n terms

72.  Given: an = n(n + 1) (n + 4) = n3 + 5n2 + 4n

73.  Given: an = n2 + 2n

74.  Given: an = (2n – 1)= 4n2 – 4n + 1

75.  Here, am+n = a + (m + n – 1)d    …(i)

and am-n = a + (m + n – 1)d    …(i)

To prove: am+n + am-n = 2am

Adding eq. (i) and (ii), we get

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

⇒ am+n + am-n = a + md + nd – d + a + md – nd – d

⇒ am+n + am-n = 2a + md + -2d

⇒ am+n + am-n = 2(a + md – d)

⇒ am+n + am-n = 2[a + (m – 1)d]

⇒ am+n + am-n = 2am

76.  Let (a – d), a, (a + d) be three numbers in A.P.

According to question, (a – d) + a + (a + d) = 24

⇒ 3a = 24

⇒ a = 8

And (a – d) (a) (a + d) = 440

⇒ (a2 – b2)a = 440

⇒ (64 – d2)8 = 440

⇒ 64 – d2 = 55

⇒ d2 = 64 – 55

⇒ d2 = 9

⇒ d = ±3

Taking d = 3, A.P. is (8 – 3), 8, (8 + 3)

⇒ 5, 8, 11

Taking d = -3, A.P. is (8 + 3), 8, (8 – 3)

⇒ 11, 8, 5.

77.  

78.  Given: A.P. 203, 210, 217, …….., 399

Here a = 203, d = 210 – 203 = 7 and an = 399

Now, an = a+(n – 1)d

79.  Given: A.P. which is divisible by 2

2, 4, 6, ….., 100

Here a = 2, d = 4 – 2 = 2 and an = 100

Now, an = a+(n – 1)d

Again A.P. which is divisible by 5

5, 10, 15, …..100

Here a = 5, d = 10 – 5 = 5 and an = 100

Now, an = a+(n – 1)d

Again A.P. which is divisible by both 2 and 5

10, 20, 30, ….., 100

Here a = 10, d = 20 – 10 = 10 and an = 100

Now, an = a+(n – 1)d

Now sum of integers divisible by 2 or 5 = sum of integers divisible by 2 + sum of integers divisible by 5-sum of integers divisible by 2 and 5

∴ Required sum = (2550 + 1050) – 550 = 3050.

80.  Given: A.P. 13, 17, 21, ………., 97

Here a = 13, d = 17 – 13 = 4 and an = 97

Now, an = a+(n – 1)d

81.  Given: f(1) = 3 and f(x + y) = f(x) f(y) for all x, y ∈ N ….(i)

Putting x = 1, y = 1 in eq.(i), f(1 + 1) = f(1)f(1)

⇒ f(2) = 3 × 3 = 9

Putting x = 1, y = 2 in eq.(i) f(1 + 2) f(2)

⇒ f(3) = 3 × 9 = 27

Putting x = 1, y = 2 in eq.(i) f(1 + 3) f(3)

⇒ f(4) = 3 × 27 = 81

82.  Given: a = 15, r = 2 and Sn = 315

Hence the number of terms = 6 and the last term = 160.

83.  Given: a = 1 and a3 + a5 = 90

 which is not possible

Therefore, common ratio is r = ±3.

84.  Let a, ar, ar2 be three numbers in G.P. therefore a + ar + ar2 = 56

⇒ a(1 + r + r2) = 56   …(i)

According to question, a – 1, ar – 7, ar2 – 21 are in A.P.

Dividing eq.(i) by eq.(ii), 

Putting r = 2 in eq.(i), a(1 + 2 + 2) = 56

Then the required numbers are 8, 16, 32.

Putting

Then the required numbers are 32, 16, 8.

85.  Let the number of terms be 2n then we have the number of odd terms is n

Let the G.P. be a, ar, ar2, ………ar2n-1

Then the odd terms a, ar2, ar4, ar6, …….. form a G.P.

86.  Given: a = 11 and S4 = 56

Also, l + (l – d) + (l – 2d) + (l – 3d) = 112

⇒ 4l – 6d = 112

⇒ 4l = 112 + 6d

⇒ 4l = 112 + 6 × 2

⇒ 4l = 112 + 12

⇒ 4l = 124

⇒ = 31

∴ a = a + (n – 1)d

⇒ 31 = 11 + (n – 1) × 2

⇒ 2 (n – 1) = 20

⇒ n – 1 = 10

⇒ n = 11

87.  

88.  Let the G.P. be a, ar, ar2, ar3, ………, arn-1

Hence proved.

89.  According to question, a = a+(p – 1)d = a,    a = a + (q – 1)d = b and a = a+(r – 1)d = c

Now (q – r)a + (r – p)b + (p – q)c = 0

Putting values of a, b and c, we get

⇒ 0 = 0 Proved.

90.  

[Adding 1 to each term in the sequence]

[Dividing each fraction by ab + bc + ca]

 are in A.P. [Multiplying each fraction by abc]

⇒ a, b, c are in A.P.

91.  Given: a, b, c, d are in G.P.

To Prove: (an + bn), (bn + cn), (cn + dn) are in G.P.

92.  

⇒ a + ak = 3

⇒ a (1 + k) = 3         …(i)

And 

⇒ a (ak) = p

⇒ a2k = p          …(ii)

Also c, d are roots of x2 – 12x + q = 0

⇒ ak2 + ak3 = 12

⇒ ak2 (1 + k) = 12    …(iii)

And 

⇒ ak2 (ak3) = q

⇒ a2k5 = q             …(iv)

Dividing eq.(iii) by eq,(i), 

⇒ k2 = 4

⇒ k = ±2

Therefore, (q + p) : (q – p) = 17 : 15.

93.  

By componendio and dividendo,

Again by componendio and dividendo,

Therefore, 

94.  Since, a, b, c are in A.P.

Since, b, c, d are in G.P. 

Putting values of b and d in eq.(i) 

⇒ c2 = ae which shows that a, c, e are in G.P.

95.  (i) Sn = 5 + 55 + 555 + …….up to n terms

= 5 [1 + 11 + 111 + …… up to n terms]

(ii) Sn = .6 + .66 + .666 + …….up to n terms

= 6 [.1 + .11 + .111 + …… up to n terms]

96.  Given: 2 × 4 + 4 × 6 + 6 × 8 + ……+n terms

∴ an = (nth term of 2, 4, 5, …..) (nth term of 4, 6, 8, …..)

⇒ an = [2 + (n – 1)2] [4 + (n – 1)2] = 2n (2n + 2)

∴ a20 = 2 × 20 (2 × 20 + 2) = 40 × 42 = 1680.

97.  Given: Sn = 3 + 7 + 13 + 21 + 31 + ……. + an-1 + a      …(in)

Also S = 3 + 7 + 13 + 21 + 31 + ………+ an-2 + an-1 + an     …(ii)

Subtracting eq.(i) from eq.(ii), 0 = 3 + (4 + 6 + 8 + 10+  ……. up to (n – 1) terms) – an

98.  

99.  

100.  

101.  Total cost of the tractor = Rs. 12000, Cash paid = Rs. 6000

Balance to be paid = 12000 – 6000 = Rs. 6000

Annual instalment = Rs. 500

∴ Number of instalment  12

Interest of 1st instalment  Rs. 720

Amount of 1st installment = 500 + 720 = Rs. 1220

Interest of 2nd instalment Rs. 660

Amount of 2nd instalment = 500 + 660 = Rs. 1160

Interest of 3rd instalment Rs. 600

Amount of 3rd instalment = 500 + 600 = Rs. 1100

∴ Sequence of instalments is 1220, 1160, 1100, ………. which is in A.P

Here, a = 1220, d = 1160 – 1220 = -60 and n = 12

= 6 [2440 – 660] = Rs. 10680

Therefore, the total cost of tractor is (10680 + 6000) = Rs. 16680.

102.  Total cost of the scooter = Rs. 22000, Cash paid = Rs. 4000

Balance to be paid = 22000 – 4000 = Rs. 18000

Annual instalment = Rs. 1000

∴ Number of instalment =  = 18

Interest of 1st instalment =  = Rs. 1800

Interest of 1st instalment =  = Rs. 1800

Amount of 1st instalment = 1000 + 1800 = Rs. 2800

Interest of 2nd instalment =  = Rs. 1700

Amount of 2nd instalment = 1000 + 1700 = Rs. 2700

Interest of 3rd instalment = = Rs. 1600

Amount of 3rd instalment = 1000 + 1600 = Rs. 2600

∴ Sequence of instalments is 2800, 2700, 2600, ………. in A.P

Here, a = 2800, d = 2700 – 2800 = -100 and n = 18

Therefore, the total cost of tractor is (35100 + 4000) = Rs. 39100.

103.  Total letters in the first set = 4, Total letters in the second set = 42 = 16

Total letters in the third set = 43 = 64

∴ Sequence of letters is 4, 16, 64, ………. in G.P.

Hence,total number of letters mailed = 87380

The amount of postage on each letter = 50 paise

Therefore total amount spent on postage = 87380 × 0.50 = Rs. 43690.

104.  Total amount deposited = Rs. 10000, Rate of interest = 5% per annum

Interest of first year =  = Rs. 500

Here a = 10000, d = 500

Amount in 15th year = a15 = 10000 + (15 – 1) × 500 = 10000 + 7000 = Rs. 17000

Total amount after 20 years = Amount in the 21st year = a21 = 10000 + (21 – 1)500 = 10000 + 10000 = Rs.20000.

105.  Present value of the machine = Rs. 15625

Rate of depreciation = 20%

After 1 year value of machine =15625 – 15625 = 15625 – 3125 = Rs. 12500

After 2 year value of machine = 12500 – 12500 = 12500 – 2500 = Rs. 10000

After 3 year value of machine = 10000 – 10000 = 10000 – 2000 = Rs. 8000

∴ Sequence of values of machine after depreciation is 12500, 10000, 8000, ……. is a G.P.

Here,

Therefore, the value of machine at the end of 5 years in Rs. 5120.

106.  Number of workers on the first day = 150

Number of workers on the second day = 150 – 4 = 146

Number of workers on the third day = 146 – 4 = 142

∴ Sequence of number of workers is 150, 146, 142, ……… in A.P.

Here a = 150, d = 146 – 150 = -4

∴ Total number of workers required to finish the work in n days.

If no worker had dropped out, then the work should have finished in (n – 8) days with 150 workers on each day.

∴ Total number of workers required to finish the work in (n – 8) days = 150 (n – 8)  …(ii)

From eq.(i) and (ii), n(152 – 2n) = 150(n – 8)

⇒ 152n – 2n2 = 150n – 1200

⇒ 2n2 – 2n – 1200 = 0

⇒ n2 – n – 600 = 0

⇒ (n – 25) (n + 24) = 0

⇒ n = 25 and n = -24 which is not possible

Therefore, the work was completed in 25 days.

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