Multiple Choice Question (MCQ)

In the arithmetic progression series \{a_n\} , if a_1=\dfrac{1}{3} , a_2+a_5=4 , a_n=33 , what is n ?

  1. ×

    48

  2. ×

    49

  3. 50

  4. ×

    51

Collected in the board: Arithmetic sequence

Steven Zheng posted 10 months ago


Answer

  1. 50

    a_1=\dfrac{1}{3} , a_2+a_5=4

    a_2 = a_1+(2-1)d=a_1+d
    (1)
    a_5 = a_1+(5-1)d=a_1+4d
    (2)

    Subtract (1) from (2)

    a_5-a_2 = 3d
    (3)
    a_2+a_5=4
    (4)

    Subtract (3) from (4)

    a_2 = \dfrac{4-3d}{2}
    (5)

    subtact (1) from (5)

    \dfrac{4-3d}{2}=a_1+d

    \dfrac{4-3d}{2}=\dfrac{1}{3} +d

    12-9d=2+6d

    d = \dfrac{2}{3}

    a_n = a_1 +(n-1)d = 33

    = \dfrac{1}{3} + (n-1)\dfrac{2}{3} = 33

    1+2n-2 = 99

    n = 50

Steven Zheng posted 10 months ago

Scroll to Top