91. If p, q, r are in Geometric Progression, then which is true among the following ?

1. 1. q =	p + r
      	2

   
   
   

   2. p² = qr

   
   
   

   3. q = √pr

   
   
   

   4. p	=	r
      r		q

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Basic concept of G.P., p, q, r are in geometric progression.
   

   ∴	q	=	r	⇒ q² = pr
   	p		q

   
   

   Correct Option: C

   Using Basic concept of G.P., p, q, r are in geometric progression.
   

   ∴	q	=	r	⇒ q² = pr
   	p		q

   
   q = √pr

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92. Terms a, 1, b are in Arithmetic Progression and terms 1, a,b are in Geometric Progression. Find ‘a’ and ‘b’ given a ≠ b.
    

1. 1. 2,4
      

   
   
   

   2. –2, 1
      

   
   
   

   3. 4,1
      

   
   
   

   4. –2, 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   As per the given question ,
   a, 1, b are in A.P.
   

   ∴ Mean =	a + b
   	2

   
   

   ∴ 1 =	a + b
   	2

   
   ⇒ a + b = 2 ....(i)
   Again, 1, a, b are in G.P.
   ∴ a² = b ......(ii)
   ∴ a + a² = 2
   ⇒ a² + a – 2 = 0
   

   Correct Option: D

   As per the given question ,
   a, 1, b are in A.P.
   

   ∴ Mean =	a + b
   	2

   
   

   ∴ 1 =	a + b
   	2

   
   ⇒ a + b = 2 ....(i)
   Again, 1, a, b are in G.P.
   ∴ a² = b ......(ii)
   ∴ a + a² = 2
   ⇒ a² + a – 2 = 0
   ⇒ a² + 2a – a – 2 = 0
   ⇒ a (a + 2) – 1 (a + 2) = 0
   ⇒ (a – 1) (a + 2) = 0
   ⇒ a = –2 , 1 and b = 4 , 1
   ∴ b = 4 since a ≠ b

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93. The fifth term of the sequence for which t₁ = 1, t₂ = 2 and t_{n + 2} = t_n + t_{n + 1}, is
    

1. 1. 5
      

   
   
   

   2. 10
      

   
   
   

   3. 6
      

   
   
   

   4. 8

   
   
   

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1. View Hint View Answer Discuss in Forum

   From the question ,
   t_{n + 2} = t_n + t_{n + 1}
   Putting on n = 1 , 2 and 3 respectively , we get
   t₃ = t₁ + t₂ = 3
   t₄ = t₃ + t₂ = 3 + 2 = 5
   

   Correct Option: D

   From the question ,
   t_{n + 2} = t_n + t_{n + 1}
   Putting on n = 1 , 2 and 3 respectively , we get
   t₃ = t₁ + t₂ = 3
   t₄ = t₃ + t₂ = 3 + 2 = 5
   ∴ t₅ = t₄ + t₃ = 3 + 5 = 8

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94. 1 + (3 + 1) (3² + 1) (3⁴ + 1) (3⁸ + 1) (3¹⁶ + 1) (3³² + 1) is equal to
    

1. 1. 364 - 1

      2

   
   
   

   2. 364 + 1

      2

   
   
   

   3. 3⁶⁴ – 1

   
   
   

   4. 3⁶⁴ + 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to question ,
   

   1 + (3 + 1)(3² + 1)(34 + 1)(38 +1)(316 + 1)(332 + 1) = 1 +	(3 - 1)(3 + 1)	(3² + 1)(34 + 1)......(332 + 1)
   	3 - 1

   
   

   = 1 +	(3² - 1)(3² + 1)(34 + 1)......(332 + 1)
   	2

   
   

   = 1 +	(34 - 1)(34 + 1)(38 + 1)......(332 + 1)
   	2

   
   

   = 1 +	(38 - 1)(38 + 1)(316 + 1)(332 + 1)
   	2

   
   

   = 1 +	(316 - 1)(316 + 1)(332 + 1)
   	2

   
   

   = 1 +	(332 - 1)(332 + 1)
   	2

   
   

   Correct Option: B

   According to question ,
   

   1 + (3 + 1)(3² + 1)(34 + 1)(38 +1)(316 + 1)(332 + 1) = 1 +	(3 - 1)(3 + 1)	(3² + 1)(34 + 1)......(332 + 1)
   	3 - 1

   
   

   = 1 +	(3² - 1)(3² + 1)(34 + 1)......(332 + 1)
   	2

   
   

   = 1 +	(34 - 1)(34 + 1)(38 + 1)......(332 + 1)
   	2

   
   

   = 1 +	(38 - 1)(38 + 1)(316 + 1)(332 + 1)
   	2

   
   

   = 1 +	(316 - 1)(316 + 1)(332 + 1)
   	2

   
   

   = 1 +	(332 - 1)(332 + 1)
   	2

   
   

   Required answer = 1 +	364 - 1	=	364 + 1
   	2		2

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95. The sum ‘5 + 6 + 7 + 8 + .... + 19’ is equal to :
    

1. 1. 150
      

   
   
   

   2. 170
      

   
   
   

   3. 180
      

   
   
   

   4. 190

   
   
   

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1. View Hint View Answer Discuss in Forum

   1 + 2 + 3 +.... + n =	n(n + 1)
   	2

   
   ∴ 5 + 6 + 7 + .... + 19 = (1 + 2 + 3 + .... + 19) – (1 + 2 + 3 + 4)
   

   Correct Option: C

   1 + 2 + 3 +.... + n =	n(n + 1)
   	2

   
   ∴ 5 + 6 + 7 + .... + 19 = (1 + 2 + 3 + .... + 19) – (1 + 2 + 3 + 4)
   

   Required answer =	19(19 + 1)	- 10 = 180
   	2

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