Sequences and Series

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  1. In a GP, the first term is 5 and the common ratio is 2. The eighth term is —








  1. View Hint View Answer Discuss in Forum

    Here , first term a = 5 and the common ratio r = 2. T
    According to question,
    nth term of a GP = arn – 1
    ∴ 8th term = 5 × (2)8 – 1 = 5 × (2)7

    Correct Option: A

    Here , first term a = 5 and the common ratio r = 2. T
    According to question,
    nth term of a GP = arn – 1
    ∴ 8th term = 5 × (2)8 – 1 = 5 × (2)7
    8th term = 5 × 128 = 640
    Hence , The eighth term is 640 .

  1. If the arithmetic mean of two numbers is 5 and geometric mean is 4, then the numbers are —








  1. View Hint View Answer Discuss in Forum

    Given in question , arithmetic mean of two numbers is 5 and geometric mean is 4.
    Let the two numbers be a and b. Then,

    Arithmetic mean =
    a + b
    		 = 5
    2

    ⇒ a + b = 10
    and Geometric mean = √ab = 4
    On squaring in ( 1) both sides , we get
    ab = 16 ...(i)
    We know that , (a – b)² = (a + b)² – 4ab
    ⇒ (a – b)² = 100 – 64 = 36

    Correct Option: D

    Given in question , arithmetic mean of two numbers is 5 and geometric mean is 4.
    Let the two numbers be a and b. Then,

    Arithmetic mean =
    a + b
    		 = 5
    2

    ⇒ a + b = 10
    and Geometric mean = √ab = 4
    On squaring in ( 1) both sides , we get
    ab = 16 ...(i)
    We know that , (a – b)² = (a + b)² – 4ab
    ⇒ (a – b)² = 100 – 64 = 36
    ⇒ a – b = 6 ...(ii)
    Solving Eqs. (i) and (ii) ,we get
    Hence , a = 8 and b = 2

  1. What is the next number in the series given below ?
    2, 5, 9, 14, 20








  1. View Hint View Answer Discuss in Forum

    According to question,
    Adding +3 , +4 , +5 , +6 , +7 , we get

    Correct Option: C

    According to question,
    Adding +3 , +4 , +5 , +6 , +7 , we get

    Hence, the next number of the series will be 27.

  1. The sum of 40 terms of an AP whose first term is 4 and common difference is 4, will be —








  1. View Hint View Answer Discuss in Forum

    Given that , In A.P. , first term a = 4 and common difference r = 4 , n = 40
    Using the given formula ,

    S40 =
    n
    		[2a + (n - 1)d]
    2

    Correct Option: A

    Given that , In A.P. , first term a = 4 and common difference r = 4 , n = 40
    Using the given formula ,

    S40 =
    n
    		[2a + (n - 1)d]
    2

    S40 = 20 [4 + 39 × 4]
    ∴ S40 = 20 × 160 = 3200

  1. Let Sn denote the sum of the first ‘n’ terms of an AP
    S2n = 3Sn Then, the ratio
    S3n
    		is equal to
    Sn









  1. View Hint View Answer Discuss in Forum

    Let a be the first term and d be the common difference.

    Then, Sn =
    n
    		[2a + (n - 1)d]
    2

    S2n =
    2n
    		[2a + (2n - 1)d]
    2

    and S3n =
    3n
    		[2a + (3n - 1)d]
    2

    Given, S2n = 3Sn
    ∴
    2n
    		[2a + (2n - 1)d] = 3
    n
    		[2a + (n - 1)d]
    22

    ⇒ 4a + (4n – 2)d = 6a + (3n – 3)d
    ⇒ d (4n – 2 – 3n + 3) = 2a
    ⇒ d =
    2a
    n + 1

    ∴ Sn =
    2an²
    n + 1

    and S3n =
    12an²
    n + 1

    Correct Option: B

    Let a be the first term and d be the common difference.

    Then, Sn =
    n
    		[2a + (n - 1)d]
    2

    S2n =
    2n
    		[2a + (2n - 1)d]
    2

    and S3n =
    3n
    		[2a + (3n - 1)d]
    2

    Given, S2n = 3Sn
    ∴
    2n
    		[2a + (2n - 1)d] = 3
    n
    		[2a + (n - 1)d]
    22

    ⇒ 4a + (4n – 2)d = 6a + (3n – 3)d
    ⇒ d (4n – 2 – 3n + 3) = 2a
    ⇒ d =
    2a
    n + 1

    ∴ Sn =
    2an²
    n + 1

    and S3n =
    12an²
    n + 1

    ∴
    Sn
    		 =
    2an²
    		 ×
    n + 1
    		 =
    1
    S3nn + 112an²6

    ⇒
    S3n
    		 = 6
    Sn