Math

If 6, p and 14 are consecutive terms in arithmetic progression(A.P) Find the value of p

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  1. (14 - 6) / 2 = p

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    Ms. Sue
  2. I think she meant p = (6+14)/2 = 10
    the middle term is the average of the other two.

    This is because the difference is constant. That is, p-6 = 14-p

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  3. Oops! You're right, oobleck!

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    Ms. Sue
  4. Let first term= a
    second term= T2
    third term= T3
    nth term= Tn
    common differnce= d
    So, a= 6
    T2= P
    T3= 14
    d= ?
    Tn= a + (n - 1)d
    T2= a + (2 - 1)d= P
    = a + d= P
    = 6 + d= P
    T3= a + (3 - 1)d= 14
    = a + 2d= 14
    = 6 + 2d= 14
    = 2d= 14 - 6
    = 2d= 8
    = d= 8/2= 4
    = d= 4

    Thus, from T2,
    6 + d= P (d= 4)
    6 + 4= P
    10= P
    P= 10

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  5. Answer:10

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