# 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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