**1.** Given: a_{n} = n(n + 2)

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

a_{n} = n(n + 2) = 1 × 3 = 3

a_{n} = n(n + 2) = 2 × 4 = 8

a_{n} = n(n + 2) = 3 × 5 = 15

a_{n} = n(n + 2) = 4 × 6 = 24

a_{n} = n(n + 2) = 5 × 7 = 35

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

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

Therefore, the first five terms are

**3. ** Given: a_{n} = 2

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

a_{1} = 2^{1} = 2

a_{2} = 2^{2} = 4

a_{3} = 2^{3} = 8

a_{4} = 2^{4} = 16

a_{5} = 2^{5} = 32

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

**5. ** Given: a = (-1) .5

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

a_{1} = (-1)^{1-1} .5^{1+1} = (-1)^{0} .5^{2} = 1 × 25 = 25

a_{2} = (-1)^{2-1} 5^{2+1} = (-1)^{1} .5^{3} = -1 × 125 = -125

a_{3} = (-1)^{3-1} .5^{3+1} = (-1)^{2} .5^{4} = 1 × 625 = 625

a_{4} = (-1)^{4-1} .5^{4+1} = (-1)^{3} .5^{5} = 1 × 3125 = -3125

a_{5} = (-1)^{5-1} .5^{5+1} = (-1)^{4} .5^{6} = 1 × 15625 = 15625

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

**7.** Given: a_{n} = 4n – 3

∴ a_{17}, = 4 × 17 – 3 = 68 – 3 = 65

a_{24 }= 4 × 24 – 3 = 96 – 3 = 93

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

**9. ** Given: a_{n} = (-1)^{n-1} n^{3}

∴ a_{9}= (-1)^{9-1} × (9)^{3} = (-1)^{8} × 729 = 729

Therefore, 9th term is 729.

**11. ** a_{n} = 3, a_{n} = 3a_{n-1} + 2 for all n > 1

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

a_{2} = 3a_{2-1} + 2 = 3a_{1} + 2 = 3 × 3 + 2 = 9 + 2 = 11

a_{3} = 3a_{3-1} + 2 = 3a_{2} + 2 = 3 × 11 + 2 = 33 + 2 = 35

a_{4} = 3a_{4-1} + 2 = 3a_{3} + 2 = 3 × 35 + 2 = 105 + 2 = 107

a_{5} = 3a_{5-1} + 2 = 3a_{4} + 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 + …..

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

**13. **Given: a_{1} = a_{2} = 2, a_{n} = a_{n-1} – 1, n > 2

Putting n = 3, 4 and 5 we get,

a_{3} = a_{3-1} – 1 = a_{2} – 1 = 2 – 1 = 1

a_{4} = a_{4-1} – 1 = a_{3} – 1 = 1 – 1 = 0

a_{5} = a_{5-1} – 1 = a_{4} – 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: a_{1} = a_{2} and a_{n-1} + a_{n-1}, n > 2

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

a_{3} = a_{3-1} + a_{3-2} = a_{2} + a_{1} = 1 + 1 = 2

a_{4} = a_{4-1} + a_{4-2} = a_{3} + a_{2} = 2 + 1 = 3

a_{5} = a_{5-1} + a_{5-2} = a_{4} + a_{3} = 3 + 2 = 5

a_{6} = a_{6-1} + a_{6-2} = a_{5} + a_{4} = 5 + 3 = 8

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

Here, a = 1, d = 3 – 1 = 2 and a_{n} = 2001

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

Here a = 105, d = 110 – 105 = 5 and a_{n} = 995

**17. **According to question, a = 2 and

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

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 S_{n} = 116

Now, a_{n} = a + (n – 1)d

⇒ a_{8} = 25 + (8 – 1) × (-3)

⇒ a_{8} = 25 – 21 = 4.

**21. ** Given: a_{k} = 5k + 1

Putting k = 1 and k = n, we get

a = 5 × 1 + 1 = 6 and a_{n} = 5n + 1

**22.** Given: S_{n} = pn + qn^{2}

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

Now S_{n} = pn + qn^{2}

**23.** Let a_{1}, a_{2} and d_{1}, d_{2} 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, S_{p} = S_{q}

**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: S_{n} = 3n^{2} + 5n and a_{m} = 164

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

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

We have a = S – S

∴ a_{n} = S_{2} – S_{1}

⇒ a_{2} = 22 – 8 = 14

Now d = a_{2} – a_{1} = 14 – 8 = 6

∴ a_{m} = 164 ⇒ a + (m – 1)d = 164

⇒ 8 + (m – 1)6 = 164

⇒ 8 + 6m – 6 = 164

⇒ 6m = 162

⇒ m = 162/6 = 27

**28.** Let A_{1}, A_{2}, A_{3}, A_{4} and A_{5} be five numbers between 8 and 26 such that

8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26 are in A.P.

Here, a = 8 and a_{7} = 26 and let d be the common difference.

∴ a_{7} = a + (7 – 1)d = 26

⇒ 8 + 6d = 26

⇒ 6d = 18

⇒ d = 3

Now, A_{1} = a + d = 8 + 3 = 11

A_{2} = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A_{3} = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A_{4} = a + 4d = 8 + 4 × 3 = 8 + 12 = 20

A_{5} = 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 A_{1}, A_{2}, A_{3}, A_{4}, …….., A_{m} 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

⇒ a_{30} = 100 + (30 – 1) × 5

⇒ a_{30} = 100 + 29 × 5

⇒ a_{30} = 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 a_{16} = a + 15d = 120 + 15 × 5 = 195° > 180°

Therefore, number of sides of the polygon are 9.

**34. ** Let a be the first term of given G.P. Here r = 2 and a_{8} = 192

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

∴ a_{5} = p ⇒ ar^{4} = p ..(i)

a_{8} = q ⇒ ar^{7} = q ..(ii)

a_{11} = s ⇒ ar^{10} = s ..(ii)

Squaring both sides of eq.(ii), we get a^{2} = (ar^{7})^{2}

⇒ q^{2} = a^{2}r^{14}

⇒ q^{2} = (ar^{4}) (ar^{10})

⇒ q^{2} = 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 a_{4} = (a_{2})^{2}

Now, a_{4} = (a_{2})^{2}

⇒ ar^{3} = (ar)^{2}

⇒ ar^{3} = a^{2}r^{2}

⇒ r = a

⇒ r = -3 [∵ a = -3]

∴ a_{7} = ar^{7-1} = (-3) × (-3)^{6}

= -3 × 729 = -2187.

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

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

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

Putting the value of S_{n} in eq.(i), we get

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

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 + ar^{2} = 16

⇒ a(1 + r + r^{2}) = 16 …(i)

And ar^{3} + ar^{4} + ar^{5} = 128

⇒ ar^{3} (1 + r + r^{2}) = 128 …(ii)

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

16r^{3} = 128

⇒ r^{3} = 8

⇒ r = 2

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

**47.** Given: a = 729 a_{7} = 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 a_{5} = 4a_{3}

⇒ ar^{4} = 4ar^{2}

⇒ r^{2} = 4

⇒ r = ±2

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

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.

∴ a_{4} = x

⇒ ar^{3} = x …(i)

a_{10} = y

⇒ ar^{9} = y …(ii)

a_{16} = z

⇒ ar^{15} = z …(iii)

From eq.(ii), ar^{9} = y

⇒ (ar^{8})^{2} = y^{2}

⇒ y^{2} = (ar^{3}) (ar^{15})

⇒ y^{2} – xz [From eq.(i) and (iii)]

∴ x, y, z are in G.P.

**50.** Here S_{n} = 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, ar^{2}, ar^{3}

∴ ar^{2} = a + 9 and ar = ar^{3} + 18

Now, ar^{2} – a = 9

⇒ a(r^{2} – 1) = 9 …(i)

And ar – ar^{3} = 18

⇒ ar (1 – r^{2}) = 18

⇒ -ar (r^{2} – 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

ar^{2} = 3 × (-2)^{2} = 12

ar^{3} = 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 a_{n} = b

⇒ ar^{n-1} = b …(i)

Given: P = a, ar, ar^{2}, ar^{3}, ……..ar^{n-1}

⇒ P^{2} = a^{2n}r^{n(n-1)} = [aar^{n-1}]^{n} [Squaring both sides]

⇒ P^{2} = (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 = ar^{2} and d = ar^{3}

Now, L.H.S. = (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})

= (a^{2} + a^{2}r^{2} + a^{2}r^{4}) (a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6})

= a^{2} (1 + r^{2} + r^{4}) a^{2}r^{2} (1 + r^{2} + r^{4}) = a^{4}r^{2} (1 + r^{2} + r^{4})^{2}

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

= (a.ar + ar.ar^{2} + ar^{2}.ar^{3})^{2}

= (a^{2}r + a^{2}r^{3} + a^{2}r^{5})^{2}

= (a^{2}r)^{2} (1 + r^{2} + r^{4})^{2} = a^{4}r^{2} (1 + r^{2} + r^{4})^{2}

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

**58. ** Let G_{1} and G_{2} be two numbers between 3 and 81 such that 3, G_{1}, G_{2}, 81 are in G.P.

Let r be the common ration

Here a = 3 and a = 81

⇒ ar^{3} = 81

⇒ 3 × r^{3} = 81

⇒ r^{3} = 27

⇒ r = 3

∴ G_{1} = ar = 3 × 3 = 9

And G_{2} = ar^{2} = 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

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 × 2^{3-1} = 30 × 2^{2} = 120

And Bacteria at the end of 4th hour = 30 × 2^{5-1} = 30 × 2^{4} = 480

And Bacteria at the end of nth hour = a_{n+1} = 30 (2^{(n+1)-1}) = 30(2^{n})

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

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

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.

⇒ a + b = 16

And G.M = √ab = 5

⇒ ab = 25

Now, Quadratic equation x^{2} – (Sum of roots)x + (Product of roots) = 0

⇒ x^{2} – (a + b)x + ab = 0

⇒ x^{2} – 16x + 25 = 0

Therefore, required equation is x^{2} – 16x + 25 = 0.

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

∴ a_{n} = [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

∴ a_{n} = [nth term of 1, 2, 3, …….] [nth term of 2, 3, 4, 5, …….] [nth term of 3, 4, 5, …….]

**67. ** 3 × 1^{2} + 5 × 2^{2} + 7 × 3^{2} + ……to n terms

∴ a_{n} = [nth term of 3, 5, 7, …….] [nth term of 1, 2, 3, 4, 5, …….]^{2}

= (2n + 1) (n)^{2} = 2n^{3} + n^{2}

Then 1 = A(n + 1) + Bn

Put n = 0 then A = 1

Put n = -1 then B = -1

**69. ** Given: 5^{2} + 6^{2} + 7^{2} + …… + 20^{2}

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

∴ a_{n} = [nth term of 3, 6, 9, …….] [nth term of 8, 11, 14, ]

**71.** 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) +…..to n terms

**72.** Given: a_{n} = n(n + 1) (n + 4) = n^{3} + 5n^{2} + 4n

**73.** Given: a_{n} = n^{2} + 2^{n}

**74.** Given: a_{n} = (2n – 1)^{2 }= 4n^{2} – 4n + 1

**75. **Here, a_{m+n} = a + (m + n – 1)d …(i)

and a_{m-n} = a + (m + n – 1)d …(i)

To prove: a_{m+n} + a_{m-n} = 2a_{m}

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

a_{m+n} + a_{m-n} = a + (m + n – 1)d + a +(m – n – 1)d

⇒ a_{m+n} + a_{m-n} = a + md + nd – d + a + md – nd – d

⇒ a_{m+n} + a_{m-n} = 2a + md + -2d

⇒ a_{m+n} + a_{m-n} = 2(a + md – d)

⇒ a_{m+n} + a_{m-n} = 2[a + (m – 1)d]

⇒ a_{m+n} + a_{m-n} = 2a_{m}

**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

⇒ (a^{2} – b^{2})a = 440

⇒ (64 – d^{2})8 = 440

⇒ 64 – d^{2} = 55

⇒ d^{2} = 64 – 55

⇒ d^{2} = 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.

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

Here a = 203, d = 210 – 203 = 7 and a_{n} = 399

Now, a_{n} = 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 a_{n} = 100

Now, a_{n} = a+(n – 1)d

Again A.P. which is divisible by 5

5, 10, 15, …..100

Here a = 5, d = 10 – 5 = 5 and a_{n} = 100

Now, a_{n} = 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 a_{n} = 100

Now, a_{n} = 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 a_{n} = 97

Now, a_{n} = 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 S_{n} = 315

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

**83.** Given: a = 1 and a_{3} + a_{5} = 90

Therefore, common ratio is r = ±3.

**84.** Let a, ar, ar^{2} be three numbers in G.P. therefore a + ar + ar^{2} = 56

⇒ a(1 + r + r^{2}) = 56 …(i)

According to question, a – 1, ar – 7, ar^{2} – 21 are in A.P.

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

Then the required numbers are 8, 16, 32.

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, ar^{2}, ………ar^{2n-1}

Then the odd terms a, ar^{2}, ar^{4}, ar^{6}, …….. form a G.P.

**86.** Given: a = 11 and S_{4} = 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

**88.** Let the G.P. be a, ar, ar^{2}, ar^{3}, ………, ar^{n-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.

[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: (a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + d^{n}) are in G.P.

⇒ a + ak = 3

⇒ a (1 + k) = 3 …(i)

⇒ a (ak) = p

⇒ a^{2}k = p …(ii)

Also c, d are roots of x^{2} – 12x + q = 0

⇒ ak^{2} + ak^{3} = 12

⇒ ak^{2} (1 + k) = 12 …(iii)

⇒ ak^{2} (ak^{3}) = q

⇒ a^{2}k^{5} = q …(iv)

⇒ k^{2} = 4

⇒ k = ±2

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

By componendio and dividendo,

Again by componendio and dividendo,

**94.** Since, a, b, c are in A.P.

Putting values of b and d in eq.(i)

⇒ c^{2} = ae which shows that a, c, e are in G.P.

**95.** (i) S_{n} = 5 + 55 + 555 + …….up to n terms

= 5 [1 + 11 + 111 + …… up to n terms]

(ii) S_{n} = .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

∴ a_{n} = (nth term of 2, 4, 5, …..) (nth term of 4, 6, 8, …..)

⇒ a_{n} = [2 + (n – 1)2] [4 + (n – 1)2] = 2n (2n + 2)

∴ a_{20} = 2 × 20 (2 × 20 + 2) = 40 × 42 = 1680.

**97.** Given: S_{n} = 3 + 7 + 13 + 21 + 31 + ……. + a_{n-1} + a …(i_{n})

Also S = 3 + 7 + 13 + 21 + 31 + ………+ a_{n-2} + a_{n-1} + a_{n} …(ii)

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

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

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

Annual instalment = Rs. 500

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

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 = 4^{2} = 16

Total letters in the third set = 4^{3} = 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 = a_{15} = 10000 + (15 – 1) × 500 = 10000 + 7000 = Rs. 17000

Total amount after 20 years = Amount in the 21st year = a_{21} = 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 – 2n^{2} = 150n – 1200

⇒ 2n^{2} – 2n – 1200 = 0

⇒ n^{2} – 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.