arithmetic progression

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the 9th term of an AP is 52 while the 12th term is 70. find the sum of it's 20th term?

  • arithmetic progression -

    9th term = 52 ---> a + 8d = 52
    12th term = 70 --> a + 11d = 70
    subtract them
    3d = 18
    d = 6
    then a+48 = 52
    a = 4

    20th term = a + 19d = 4 + 114 = 118

  • arithmetic progression -

    118

  • arithmetic progression -

    118

  • arithmetic progression -

    9th term = 52
    a+8d=52
    12th term = 70
    a+11d =70
    Subtract them
    a+8d=52
    -
    a+11d =70
    3d=18
    d=6 buy dividing both side by 3
    To fine a
    a+8d=52
    When d= 6
    a+8(6)=52
    a+48=52
    a=52-48
    a=4
    Then the sum of 20th term
    Will be
    S20= n/2[2a + (n-1)d]
    S20= 20/2[2(4) + (20-1)6]
    S20=10[8+19(6)]
    S20=10[8+114]
    S20=10[122]
    S20=1220 answer

  • arithmetic progression -

    9th term = 52
    a+8d=52
    12th term = 70
    a+11d =70
    Subtract them
    a+8d=52
    -
    a+11d =70
    3d=18
    d=6 buy dividing both side by 3
    To fine a
    a+8d=52
    When d= 6
    a+8(6)=52
    a+48=52
    a=52-48
    a=4
    Then the sum of 20th term
    Will be
    S20= n/2[2a + (n-1)d]
    S20= 20/2[2(4) + (20-1)6]
    S20=10[8+19(6)]
    S20=10[8+114]
    S20=10[122]
    S20=1220 answer

  • arithmetic progression -

    Since is sum of 20th term not the 20th term then the answer is 1220

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