Arithmetic Progression

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  • Q1

    If a, b, c are in A.P., then prove that ka, kb, kc are in A.P.

    Marks:2
    Answer:

    If a, b, c are in A.P. then

    b-a = c-b ...(i)

    Now, ka, kb and kc will be in A.P.

    If kb- ka = kc- kb

    or k(b-a) = k(c-b)

    or b-a = c-b which is true by (i) .

    Hence if a, b, c are in A.P., then ka, kb, kc are also in A.P.

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  • Q2

    If 20th term of an A.P. is 99 and 3rd term is 14, find its first and 4th term.

    Marks:2
    Answer:

    Tn = a + (n - 1)d

    T
    20 = 99 = a + (20 - 1)d
    a + 19d= 99 ...(i)

    T3 = 14 = a + (3 - 1)d
    a + 2d = 14 ...(ii)

    Solving these two equations we get, a = 4 and d = 5.

    Thus the first term is 4 and T4 = 4 + (4 - 1)5
    = 19.

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  • Q3

    Find the 12th term of an A.P. whose first term is 2 and common difference is 4.

    Marks:1
    Answer:

    For an A.P.

    Tn = a + (n - 1)d

    Here a = 2, d = 4 and n = 12.

    T12 = 2 + (12 - 1)4 = 2 + 44 = 46.

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  • Q4

    The nth term of a sequence is given by an=n2+3n-5. Find its 21st term.

    Marks:2
    Answer:

    We have,

    an=n2+3n-5

    On putting n=21, we get

    21st term=212+3x21-5

    = 441+63-5

    = 504-5

    = 499

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  • Q5

    Marks:2
    Answer:

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