Question

if the number (3p+4) (7P+1)and (12P-5) are in A.P. then the value of is

Answer 1

Hope u like my process
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IF X, Y, Z are in AP then we can say that

=> Y-X=Z-Y
.
=> 2Y = Z + X
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Now....

=> X = 3P+4

=> Y = 7P +1
.
=> Z = 12P - 5

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So,

$= > \blue{2(7p + 1)} = \blue{(3p + 4) + (12p - 5)} \\ \\ = > \blue{ 14p + 2} = \blue{15p - 1 }\\ \\ = > \blue{ 15p - 14p }= \blue{2 + 1} \\ \\ = > \boxed{p = \underline{ \bf \green{3}}}$
So the required value of P is 3
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Hope this is ur required answer

Proud to help you

Answer 2

Answer:

The value of p is 3.

In an A.P.  if a,b, and c are three consecutive terms of an A.P. then we have

$b = \frac{a+c}{2}$

Step-by-step explanation:

Given, the terms of the A.P. are  (3p+4) (7P+1), and (12P-5)

Now we have

                      $2 * (7P+1) = (3P + 4) + (12P - 5)$

                      $14P + 2 = 15P - 1$

                      $P = 3$