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**NCERT Class 11 Mathematics Chapter 8 Sequences and Series**

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**Solutions****NCERT Class 11 Mathematics Chapter 8 Sequences and Series****Sequences and Series**

**Sequences and Series****Chapter – 8**

Exercise 8.1 |

** Write the first five terms of each of the sequences in Exercises 1 to 6 whose nᵗʰterms are:**

**1. aₙ = n(n + 2).**

Ans: aⁿ= n(n + 2)

Substituting n = 1, 2, 3, 4 and 5, we obtain

a₁= 1(1 + 2) = 3

a₂ = 2(2 + 2) = 8

a₃ = 3(3 + 2) = 15

a₄ = 4(4 + 2) = 24

a₅ = 5(5 + 2) = 35

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

Ans:

**3. aₙ = 2ⁿ.**

Ans: aₙ = 2ⁿ

Substituting n = 1, 2, 3, 4, 5 we obtain

a₁ = 2¹ = 2

a₂ = 2² = 4

a₃ = 2³ = 8

a₄ = 2⁴ = 16

a⁵ = 2⁵ = 32

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

Ans:

**5. aₙ = (- 1)ⁿ⁻¹ 5ⁿ⁺¹.**

Ans: Substituting n = 1, 2, 3, 4, 5 we obtain

a₁= (- 1)¹⁻¹ 5¹⁺¹ = 5² = 25

a₂ = (- 1)²⁻¹ 5²⁺¹ = – 5³= – 125

a₃ = (- 1)³⁻¹ 5³⁺¹ = 5⁴= 625

a₄ = (- 1)⁴⁻¹ 5⁴⁺¹ = – 5⁵ = – 3125

a⁵ = (- 1)⁵⁻¹ 5⁵⁺¹= 5⁶ = 15625

Therefore, the required terms are 25, -125, 625, – 3125, and 15625.

Ans:

**Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nᵗʰ terms are: **

**7. aₙ = 4n – 3 a₁₇, a₂₄.**

Ans: Substituting n = 17 we obtain

a₁₇ = 4(17) – 3 = 68 – 3 = 65

Substituting n = 24 we obtain

a₂₄= 4(24) – 3 = 96 – 3 = 93

Ans: Substituting n = 7 , we obtain

**9. aₙ = (- 1)ⁿ⁻¹ n³ ; a₉.**

Ans: Substituting n = 9 , we obtain

a₉ = (- 1)⁹⁻¹ (9)³ = (9)³ = 729

Ans: Substituting n = 20 we obtain

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

Ans

:

⇒ a₂ = 3a₁+ 2 = 3(3) + 2 = 11

a₃ = 3a₂ + 2 = 3(11) + 2 = 35

a₄ = 3a₃ + 2 = 3(35) + 2 = 107

a₅ = 3a₄ + 2 = 3(107) + 2 = 323

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

The corresponding series is 3 + 11 + 35 + 107 + 323 +…

Ans:

Ans:

⇒ a₃ = a₂ – 1 = 2 – 1 = 1

a₄ = a₃ – 1 = 1 – 1 = 0

a₅ = a₄ – 1 = 0 – 1 = – 1

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

The corresponding series is 2 + 2 + 1 + 0 +(- 1)+…

**14. The Fibonacci sequence is defined by**

Ans: 1 = a₁ =a₂

∴ a₃ = a₂ + a₁ = 1 + 1 = 2

a₄ = a₃ + a₂ = 2 + 1 = 3

a₅ = a₄ + a₃ = 3 + 2 = 5

a₆ = a₅ + a₄ = 5 + 3 = 8

Exercise 8.2 |

Ans:

**2. Find the 12ᵗʰ term of a G.P. whose 8ᵗʰ term is 192 and the common ratio is 2.**

Ans: Common ratio, r = 2

Let a be the first term of the G.P.

∴ a₈ = ar⁸⁻¹ = ar⁷

⇒ar⁷ = 192

a(2)⁷ = 192

a(2)⁷ = (2)⁶ (3)

**3. The 5ᵗʰ, 8ᵗʰ and 1ᵗʰ terms of a G.P. are p, q and s, respectively. Show that q²= ps.**

Ans: Let a be the first term and r be the common ratio of the G.P.

According to the given condition, a₅ = ar⁵⁻¹ = a r⁴ = p … (1)

a₈ = ar⁸⁻¹ = ar⁷ = q …(2) a₁₁ = ar¹¹⁻¹ = ar¹⁰ = s…(3)

Dividing equation (2) by (1), we obtain

Dividing equation (3) by (2), we obtain

Equating the values of r³ obtained in (4) and (5), we obtain

**4. The 4ᵗʰ term of a G.P. is square of its second term, and the first term is – 3.**

**Determine its 7ᵗʰ term.**

Ans: Let a be the first term and r be the common ratio of the G.P.

∴ a = – 3

It is known that, aₙ = arⁿ⁻¹

a₄ = ar³ = (- 3) r³

a₂ = a r¹ = (- 3) r

According to the given condition,

(-3) r³ =[(-3)r] ⇒-3r³ =9 r² ⇒ r = – 3 a₇ = ar

⁷⁻¹ = a r⁶ = (- 3) (- 3)⁶ = – (3)⁷ = – 2187 Thus, the seventh term of the G.P. is -2187.

**5. Which term of the following sequences:**

**(a) 2, 2√2, 4….. is 128?**

Ans:

Thus, the 13ᵗʰ term of the given sequence is 128.

**(b) √3,3,3√3,… is729?**

Ans: The given sequence is √3 3, 3√3….

Here,

Let the nᵗʰ term of the given sequence be 729.

⇒ n = 12

Thus, the 12th term of the given sequence is 729.

Ans:

**Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:**

Ans:

**7. 0.15, 0.015, 0.0015, … 20 terms. **

Ans:

**8. √7, √21, 3√7,…n terms.**

Ans: The given G.P. is √7, √21, 3√7,…

Here, a = √7

**9. 1, – a, a ², – a³ ,…n terms (if a ≠ -1).**

Ans: The given G.P. is 1,-a, a², -a³,……….

Here, first term = a₁ = 1

Common ratio = r = – a

**10. x ³, x⁵, x⁷ ,…n terms (if x ne ≠ ± 1 ).**

Ans: The given G.P. is x³,x⁵,x⁷,…

Here, a = x³ and r = x²

Ans:

**12: The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.**

Ans:

From (2), we obtain

a³ = 1

⇒ a = 1 (Considering real roots only)

Substituting a = 1 in equation (1), we obtain

⇒ 10 + 10r + 10r² – 39r = 0

⇒ 10r² – 29r + 10 = 0

⇒ 10r² – 25r – 4r + 10 = 0

⇒ 5r(2r – 5) – 2(2r – 5) = 0

⇒ (5r – 2)(2r – 5) = 0

**13. How many terms of G.P. 3, 3²,3³,….. are needed to give the sum 120?**

Ans: The given G.P. is 3, 3²,3³,…..

Let n terms of this G.P. be required to obtain the sum as 120.

Here, a = 3 and r = 3

⇒3ⁿ – 1 = 80

⇒ 3ⁿ = 81

⇒3ⁿ = 3⁴

∴ n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.

**14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the GP.**

Ans: Let the G.P. be a, ar, ar²,ar³,…

According to the given condition, a + ar

+ ar² = 16 and ar³ + ar⁴ + ar⁵ =128 ⇒ a

(1 + r + r²) = 16 (1)

ar⁶ (1 + r + r²) = 128…… (2)

Dividing equation (2) by (1), we obtain

⇒r³ = 8

∴ r = 2

Substituting r = 2 in (1), we obtain

a(1 + 2 + 4) = 16

⇒ a(7) = 16

**15. Given a G.P. with a = 729 and 7ᵗʰ term 64, determine S₇.**

Ans: a = 729

a₇ = 64

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

It is known that, aₙ = ar ⁿ⁻¹

a₇ = ar⁷⁻¹ = (729)r⁶

64 = 729 r⁶

= (3)⁷ – (2)⁷

=2187-128

= 2059

**16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.**

Ans: Let a be the first term and r be the common ratio of the G.P.

According to the given conditions,

a₅ = 4 x a₃ ar4

= 4ar² ⇒ r² =

4

∴ r = ± 2

From (1), we obtain

Thus, the required G .P. is

4, -8, 16, -32, …

**17. If the 4ᵗʰ, 10ᵗʰ and 16ᵗʰ terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.**

Ans: Let a be the first term and r be the common ratio of the G.P.

According to the given condition, a₄ = a r³ = x (1) a₁₀ = a r⁹

=y… (2) a₁₆ = a r¹⁵ =z…(3)

Dividing (2) by (1), we obtain

Dividing (3) by (2), we obtain

Thus, x, y, z are in G. P.

**18. Find the sum to n terms of the sequence, 8, 88, 888, 8888.**

Ans: The given sequence is 8, 88, 888, 8888…

This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as

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

**19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,1/2**

Ans:

**20. Show that the products of the corresponding terms of the sequences a, ar, ar²,…arⁿ⁻¹ and A, AR, AR², ARⁿ⁻¹ form a G.P, and find the common ratio.**

Ans: It has to be proved that the sequence, aA, arAR, ar²AR², …arⁿ⁻¹ ARⁿ⁻¹, forms a G.P.

Thus, the above sequence forms a G.P. and the common ratio is rR.

**21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4ᵗʰ by 18.**

Ans: Let a be the first term and r be the common ratio of the G.P.

a₁ = a, a₂ = ar, a₃ = ar² a₄ =

ar³ By the given condition, a³ =

a₁ +9 ⇒ ar² =a+9…(1)

a₂ =a₄ +18 ⇒ ar = ar³ + 18

……(2) From (1) and (2), we

obtain a(r² -1) =9…(3)

ar (1 – r²)=18 … (4)

Dividing (4) by (3), we obtain

⇒ – r = 2

⇒ r = – 2

Substituting the value of r in (1), we obtain

4a = a + 9

⇒ 3a = 9∴

a = 3

Thus, the first four numbers of the G.P. are 3, 3 (- 2) 3 ( – 2)², and 3 (- 2)³ i.e., 3¸- 6, 12, and – 24.

**22. If the pᵗʰ, qᵗʰ and rᵗʰ terms of a G.P. are a, b and c, respectively. Prove that **

Ans: Let A be the first term and R be the common ratio of the G.P.

According to the given information,

=A⁰ × R⁰

= 1

Thus, the given result is proved.

**23. If the first and the nᵗʰ term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ. **

Ans: The first term of the G.P is a and the last term is b.

Therefore, the G.P. is a, ar, ar² ar³, …..arⁿ⁻¹, where r is the common ratio.

b= arⁿ⁻¹…(1)

P = Product of n terms

= (a) (ar) (ar²) … (arⁿ⁻¹)

= (a × a ×…a)(r × r² ×… rⁿ⁻¹)

= aₙ r ₁⁺₂⁺… (n – 1) …(2)

Here, 1, 2, …(n – 1) is an A.P.

∴ 1+2+……….+(n – 1 )

= (ab)ⁿ [Using (1)]

Thus, the given result is proved.

**24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from**

Ans: Let a be the first term and r be the common ratio of the G.P.

Since there are n terms from (n + 1)ᵗʰ to (2n)ᵗʰ term,

Sum of terms from (n + 1)ᵗʰ to (2n)ᵗʰ term

aⁿ⁺¹ = arⁿ⁺¹ ⁻¹ = arⁿ

Thus, required ratio =

Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)ᵗʰ to (2n)ᵗʰ term is 1/rⁿ.

**25. If a, b, c and d are in G.P. show that **

(a² + b² + c²)(b²+ c² + d²) = (ab + bc + cd)²

Ans: a, b, c, d

are in G.P.

Therefore, bc = ad

….(1) b² = ac … (2)

c² = bd (3)

It has to be proved that,

(a² + b² + c²)(b²+ c² + d²) =(ab+bc + cd)²

R.H.S.

= (ab + bc + cd)²

= (ab + ad + cd)² [Using (1)]

= [ab + d(a + c)]²

= a²b² + 2abd (a + c) + d² (a + c)²

= a²b² +2a²bd + 2acbd + d²(a² + 2ac + c²)

= a²b² + 2a²c² + 2b²c² + d²a² + 2d²b² + d²c²[Using (1) and (2)]

= a² b² + a²c² + a² c² + b²c² + b²c² + d²a² + d²b² + d² b² + d² c²

= a² b² + a²c² + a² d² + b² x b² + b²c² + b²d²+ c²b² + c² x c² + c² d²

[Using (2) and (3) and rearranging terms]

= a² (b²+ c² + d²) + b²(b² + c² + d²) + c² (b²+ c² + d²)

= (a²+ b²+ c²)(b² + c² + d²) =

L.H.S.

∴ L .H.S.=R.H.S

(a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)²

**26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.**

Ans: Let G₁ and G₂ be two numbers between 3 and 81 such that the series, 3, G₁, G₂, 81, forms a G.P.

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

∴ = 81 = (3) (r)³

⇒ r³ = 27

∴ r = 3 (Taking real roots only)

For r = 3,

G₁ = ar = (3)(3) = 9

G₂ = ar² = (3) (3)² = 27

Thus, the required two numbers are 9 and 27.

**27. Find the value of n so that: **

may be the geometric mean between a and b.

Ans: G. M. of a and b is √ab

By the given condition,

Squaring both sides, we obtain

⇒ a²ⁿ⁺² + 2aⁿ⁺¹ bⁿ⁺¹ + b²ⁿ⁺² = (ab)(a²ⁿ + 2aⁿbⁿ+ b²ⁿ)

⇒ a²ⁿ⁺² + 2aⁿ⁺¹ bⁿ⁺¹ + b²ⁿ⁺² = a²ⁿ⁺¹ b + 2aⁿ⁺¹ bⁿ⁺¹ + ab ²ⁿ⁺¹

⇒ a²ⁿ⁺² + b²ⁿ⁺² = a²ⁿ⁺¹b + ab²ⁿ⁺¹

⇒ a²ⁿ⁺² – a²ⁿ⁺¹ b = ab²ⁿ⁺¹ – b²ⁿ⁺²

⇒ a²ⁿ⁺¹ (a – b) = b²ⁿ⁺¹ (a – b)

**28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio **

Ans: Let the two numbers be a and b.

G.M.= √ab

According to the given condition,

a + b = 6 √ab …(1)

⇒ (a + b)² = 36(ab)

Also,

(a – b)² = (a + b)² – 4ab = 36ab – 4ab = 32ab

⇒ a – b = √32 √ab

= 4√2 √ab …(2)

Adding (1) and (2), we obtain

2a = (6 + 4√2) √ab

⇒a = (3 + 2√2) √ab

Substituting the value of a in (1), we obtain

b = 6√ab – (3 + 2√2) √ab

⇒ b = (3 – 2√2) √ab

Thus, the required ratio is (3 + 2√2):(3-2√2).

**29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A± √(A+G)(A-G).**

Ans: Given A and G are AM and GM between two numbers. Let the two number be

⇒ a + b = 2A (1)

⇒ ab = G² (2)

⇒ (a – b)² = (a + b)² – 4ab

⇒ (a – b)² = (2A)² – 4G²

⇒ (a – b)² = 4(A² – G²)

⇒ (a – b) = ± 2 √A² – G²) (3)

Taking (1) and (3)

⇒ a – b = 2√A² – G²)

⇒ a + b = 2A

Adding both 2a = 2A + 2√A² – G²)

⇒ a = A + √A² – G²)

Putting this value of a in equation (1)

⇒ a + b = 2A

⇒ A + √A² – G²) + b = 2A

⇒ b = A – √A² – G²) = A – √(A + G)(A – G)) .

Hence, the numbers are A ± √A² – G²

**30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2ⁿᵈ hour, 4ᵗʰ hour and nᵗʰ hour?**

Ans: It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P.

Here, a = 30 and r = 2 ∴

a₃ = ar² = (30) (2)² = 120

Therefore, the number of bacteria at the end of 2ⁿᵈ hour will be 120.

a₅ = ar⁴ = (30) (2)⁴ = 480

The number of bacteria at the end of 4ᵗʰ hour will be 480.

Thus, number of bacteria at the end of nᵗʰ hour will be 30 (2)ⁿ.

**31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?**

Ans: The amount deposited in the bank is Rs 500.

At the end of first year, amount =

= Rs 500 (1.1)

At the end of 2ⁿᵈ year, amount = Rs 500 (1.1) (1.1)

At the end of 3ʳᵈ year, amount = Rs 500 (1.1) (1.1) (1.1) and so on

∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)¹⁰.

**32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.**

Ans: Let the root of the quadratic equation be a and b.

According to the given condition,

Α.Μ. = a+b/2 =8⇒a+b =16 …(1)

G.M. = √ab = 5⇒ ab = 25 …(2)

The quadratic equation is given by, x² – x

(Sum of roots) + (Product of roots) = 0 x² –

x(a + b) + (ab) =0 x² – 16x+25=0 [Using

(1) and (2)]

Thus, the required quadratic equation is x² – 16x + 25 = 0

Miscellaneous Exercise on Chapter 8 |

**1. If ƒ is a function satisfying ƒ (x+y)=ƒ(x)ƒ(y) for all x, y ∈ N such that **

**find the value of n. **

Ans: It is given that, f(x + y) = f(x) x × f(y) for

all x, y ∈ N… (1) f(1) = 3

Taking x = y = 1 in (1), we obtain f (1

+1)= f(2) = f(1) f(1) = 3 × 3 = 9

Similarly,

f(1 + 1 + 1) = f(3) = f(1 + 2) = f(1) f(2) = 3 × 9 =

27f(4) = f(1 + 3) = f(1) f(3) = 3 × 27 =81∴ f(1), f (2),

f(3) ,…, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.

It is known

that,

It is given

that,

⇒ 3ⁿ – 1 = 80

⇒ 3ⁿ = 81 = 3⁴

∴ n = 4

Thus, the value of n is 4.

**2. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.**

Ans: Let the sum of n terms of the G.P. be 315.

It is known that,

It is given that the first term a is 5 and common ratio r is 2.

⇒ 2ⁿ -1 =63

⇒ 2ⁿ = 64 = (2)⁶

⇒ n=6

∴ Last term of the G.P = 6ᵗʰ term = ar⁶⁻¹ = (5)(2)⁵ = (5)(32) = 160

Thus, the last term of the G.P. is 160.

**3. The first term of a GP. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.**

Ans: Let a and r be the first term and the common ratio of the G.P.

respectively. ∴ a = 1 a₃ = ar² = r²

a₅ = ar⁴ = r⁴

∴ r² + r⁴ = 90 ⇒ r⁴ +

r² – 90 = 0

Thus, the common ratio of the G.P. is ±3.

**4. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.**

Ans: Let the three numbers in G.P. be a, ar, and ar² .

From the given condition, a + ar + ar² = 56

⇒ a(1 + r + r²) = 56

a – 1, ar – 7 , ar² – 21 forms an A.Ρ.

∴ (ar – 7) – (a – 1)=(ar² – 21) – (ar – 7)

⇒ ar – a – 6 = ar² – ar – 14

⇒ ar² – 2ar + a =8

⇒ ar² – ar – ar +a =8

⇒ a(r² +1 – 2r)=8

⇒ ar (r – 1)² =8 …(2)

⇒7(r² – 2r+1)= 1 + r + r²

⇒7r² – 14 r+7 – 1 – r – r² = 0

⇒ 6r² – 15r + 6 = 0

⇒ 6r² – 12r – 3r + 6 = 0

⇒ 6r (r – 2) – 3(r – 2)=0

when r = 2, a = 8

When

⇒ (6r – 3)(r – 2) = 0

Therefore, when r = 2 the three numbers in G.P. are 8, 16, and 32.

when r = 1/2 , the three numbers in G.P. are 32, 16, and 8.

Thus, in either case, the three required numbers are 8, 16, and 32.

**5. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.**

Ans: Let the G.P. be T₁, T₂, T₃, T₄, …T₂ₙ

Number of terms = 2n

According to the given condition

T₁ + T₂ + T₃ +…+T₂ₙ = 5[T₁ + T₃ +…+T ₂⁻₁ ]

⇒ T₁ +T₂ +T ₃ +…+ T₂ₙ – 5 [T₁ +T₃ +…+T₂⁻₁] =0

⇒ T₂ + T₄ +…+ T₂ₙ = 4[T₁ +T₃ +…+T₂⁻₁ ]

Let the G.P. be a, ar, ar² , ar³ …

⇒ ar = 4a

⇒ r = 4

Thus, the common ratio of the G.P. is 4.

6. If

then show that a, b, c and d are in G.P.

Ans: It is given that,

⇒ (a + bx)(b – cx) = (b + cx)(a – bx)

⇒ ab – acx + b² x – bcx² = ab – b² x + acx – bcx²

⇒2b²x = 2acx

⇒ b² = ac

⇒ b/a =c/b …..(1)

Also,

⇒ (b + cx)(c – dx) = (b – cx)(c + dx)

⇒ bc – bdx + c²x – cdx² = bc + bdx – c² x – cdx²

⇒2c² x = 2bdx

⇒ c² = bd

⇒ c/d = d/c …(2)

From (1) and (2), we obtain

b/a = c/b = d/c

Thus, a, b, c, and d are in G.P.

**7. Let S be the sum, P the product and R the sum of reciprocals of n terms in a GP. Prove that P² Rⁿ = Sⁿ.**

Ans: Let the G.P. be a, ar, ar² , ar³ , … arⁿ⁻¹…

According to the given information,

Hence, P² Rⁿ = Sⁿ

**8. If a, b, c, d are in GP, prove that (aⁿ + bⁿ),(bⁿ + cⁿ),(cⁿ + dⁿ) are in G.P.**

Ans: It is given that a, b, c,and d’ are in G.P.

∴ b² = ac … (1)

c² = bd … (2)

ad = bc (3)

It has to be proved that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P. i.e.,

(bⁿ + cⁿ)² = (aⁿ + bⁿ)(cⁿ + dⁿ)

Consider L.H.S.

(bⁿ + cⁿ)²= b²ⁿ + 2bⁿ cⁿ + c²ⁿ

=(b²)ⁿ +2bⁿ cⁿ + (c²)ⁿ

= (ac)ⁿ +2bⁿ cⁿ +(bd)ⁿ [Using (1) and (2)]

=aⁿ cⁿ +bⁿ cⁿ +bⁿ cⁿ + bⁿ dⁿ

= aⁿ cⁿ + bⁿ cⁿ + aⁿ dⁿ + bⁿ dⁿ [Using (3)]

= cⁿ (aⁿ + bⁿ) + dⁿ (aⁿ + bⁿ)

= (aⁿ + bⁿ)(cⁿ + dⁿ) =

R.H.S.

∴ (bⁿ + cⁿ)² = (aⁿ + bⁿ) (cⁿ + dⁿ)

Thus, (aⁿ + bⁿ), (bⁿ + cⁿ), and (cⁿ + dⁿ) are in G.P.

**9. If a and b are the roots of x² – 3x + p = 0 and c, d are roots of x² – 12x + q = 0 where a, b, c, d form a G.P. Prove that (q+p): (q-p) = 17:15.**

Ans: It is given that a and b are the roots of x² – 3x + p = 0

∴ a + b = 3 and ab = p… (1)

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

∴ c + d = 12 and cd = q …(2)

It is given that a, b, c, d are in G.P.

Let a = x, b = xr c = xr², d = xr³

From (1) and (2), we obtain x +

xr = 3

⇒ x(1 + r) = 3

Xr² + Xr³ = 12

⇒ xr² (1 + r) = 12

On dividing, we obtain

Case 1:

When r = 2 and x = 1 ,

ab = x² r = 2cd = x² r⁵

= 32

Case II:

When r =- 2 , x = – 3 ,

ab=x² r= – 18cd = x² r⁵

= – 288

Thus, in both the cases, we obtain (q + p) : ( q – p)=17:15

**10. The ratio of the A.M. and G.M. of two positive numbers a and b, is m: n. Show that a:b = (m + √m² -n²): (m-√m² -n²).**

Ans: Let the two numbers be a and b.

A.M = (a + b)/2 and G.M.= √ab

According to the given condition,

Using this in the identity (a – b)² = (a + b)²- 4ab, we obtain

Adding (1) and (2), we obtain

Substituting the value of a in (1), we obtain

Thus, a : b = (m + √m² -n²):(m-√m²-n²)

**11. Find the sum of the following series up to n terms:**

**(i) 5+55 +555+….**

Ans: 5 + 55 + 555 +…

Let Sₙ = 5 + 55 + 555 +…. to n terms

**(ii) .6+. 66+.666+….**

Ans: 0.6 + 0.66 + 0.666 +…

Let Sₙ =06.+ 0.66 + 0.666 +… to n terms

=6 [ 0.1 + 0.11 + 0.111 + ….to n terms]

**12. Find the 20ᵗʰ term of the series 2 x 4+4×6+6×8+ … + n terms.**

Ans: The given series is 2 × 4 + 4 × 6 + 6 × 8 +…r terms ∴

nᵗʰ term = aₙ = 2n × (2n + 2) = 4n² + 4n

a₂₀ = 4 (20)² + 4(20) = 4(400) + 80 = 1600 + 80 = 1680

Thus, the 20th term of the series is 1680.

**13. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?**

Ans: It is given that the farmer pays Rs 6000 in cash.

Therefore, unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

According to the given condition, the interest paid annually is

12% of 6000, 12% of 5500, 12% of 5000,…, 12% of 500

Thus, total interest to be paid = 12% of 6000 + 12% of 5500+12% of 5000 + …+ 12% of 500

= 12% of (6000 + 5500 + 5000 + … + 500)

= 12% of (500 + 1000 + 1500 + ….. + 6000)

Now, the series 500, 1000, 1500 … 6000 is an A.P. with both the first term and common difference equal to 500.

Let the number of terms of the A.P. be n.

∴ 6000 =500+(n – 1) 500

⇒ 1+( n – 1)=12

⇒ n = 12

∴ Sum of the A.P = 12/2 [2(500) + (12 – 1)(500)] = 6[1000 + 5500] = 6(6500) = 39000

Thus, total interest to be paid = 12% of (500+ 1000 + 1500 +….. + 6000)

= 12% of 39000 = Rs 4680

Thus, cost of tractor = (Rs 12000 + Rs 4680) = Rs 16680.

**14. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?**

Ans: It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

∴ Unpaid amount = Rs 22000 – Rs 4000 =Rs 18000

According to the given condition, the interest paid annually is

10% of 18000, 10% of 17000, 10% of 16000…10% of 1000

Thus, total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000+…+ 10% of 1000

= 10% of ( 18000 + 17000 + 16000 +…+1000)

= 10% of ( 1000 + 2000 + 3000 +…+18000)

Here, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both equal to 1000.

Let the number of terms be n.

∴ 18000=1000+(n – 1 )(1000)

⇒ n = 18

∴ Total interest paid = 10% of ( 18000 + 17000 + 16000 +…+1000)

= 10% of Rs 171000 =Rs 17100

∴ Cost of scooter = Rs 22000 + Rs 17100 = Rs 39100.

**15. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8ᵗʰ set of letter is mailed.**

Ans: The numbers of letters mailed forms a G.P.: 4, 4², 4⁸

First term = 4

Common ratio = 4

Number of terms = 8

It is known that the sum of n terms of a G.P. is given by

It is given that the cost to mail one letter is 50 paisa.

∴ Cost of mailing 87380 letters

=Rs 43690

Thus, the amount spent when 8ᵗʰ set of letter is mailed is Rs 43690.

**16. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15ᵗʰ year since he deposited the amount and also calculate the total amount after 20 years.**

Ans: It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

∴ Interest in first year

∴ Amount in 15ᵗʰ year = Rs

Rs 10000 + 14 x Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

∴ Amount after 20 years =

= Rs 10000 + 20 x Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

**17. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.**

Ans: Cost of machine = Rs 15625

Machine depreciates by 20% every year.

Therefore, its value after every year is 80% of the original cost i.e., 4/5 of the original cost.

∴ Value at the end of 5 years =

= 5 × 1024 = 5120

Thus, the value of the machine at the end of 5 years is Rs 5120.

**18. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.**

Ans: Let x be the number of days in which 150 workers finish the work.

According to the given information,

150 x= 150 + 146 + 142 +….(x+8) terms

The series 150 + 146 + 142 +….(x+8) terms is an A.P. with first term 146, common difference – 4 and number of terms as (x + 8)

⇒ 150x = (x + 8)[150 + (x + 7)(- 2)]

⇒ 150x = (x + 8)(150 – 2x – 14)

⇒ 150x = (x + 8)(136 – 2x)

⇒ 75x = (x + 8)(68 – x)

⇒ 75x = 68x – x² + 544 – 8x

⇒ x² + 75x – 60x – 544 = 0

⇒ x² + 15x – 544 = 0

⇒ x² + 32x – 17x – 544 = 0

⇒ x(x + 32) – 17(x + 32) = 0

⇒ (x – 17)(x + 32) = 0

⇒x = 17 or x = – 32

However, x cannot be negative.

∴ x = 17

Therefore, originally, the number of days in which the work was completed is 17.

Thus, required number of days = (17 + 8) = 25