Question

If p – 2, p + 2, 3p – 2 are in AP, then p is equal to

Answer 1

Let a,b, c are in AP
so 2b=a+c
Using this condition we get
2(p+2)= 3p-2+p-2
2p+4=4p-4
2p=8
P=4

Answer 2

According to the question, p - 2, p + 2 and 3p - 2 are in an AP.

Since these terms are in an AP, they have a common difference, i.e, the difference between the third and second term is equal to the difference between the second and the first term.

Here, the first term is p - 2, the second is p + 2 and the third term is 3p - 2.

⇒ Second term - First term = Third term - Second term.

⇒ p + 2 - (p - 2) = 3p - 2 - (p + 2)

⇒ p + 2 - p + 2 = 3p - 2 - p - 2

⇒ 2 + 2 = 2p - 4

⇒ 4 = 2p - 4

⇒ 4 + 4 = 2p

⇒ 8 = 2p

⇒ 8/2 = p

⇒ 4 = p

$\Large\boxed{\sf The \ value \ of \ p = 4}$

Explanation of the concept used:

When a number of terms are in an Arithmetic Progression, the terms taken two at a time have a common difference, and the common difference is denoted by 'd'.

Consider an AP: a₁, a₂, a₃, a₄, . . . . . . a$\sf{_n}$

Here, the common difference will be the difference between a term and the preceding term.

Considering the second term:

➝ Common difference = a₂ - a₁ → Eq(1)

Considering the third term:

➝ Common difference = a₃ - a₂ → Eq(2)

From Eq(1) and Eq(2):

➝ a₂ - a₁ = a₃ - a₂

We've applied the above concept in the question to find the value of 'p'.