Question

If 3, 4 + P, 6-P are in A.P. then P must be equal to


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Answer 1

Answer:-

• p = -1 and p = 1/2

Given:-

• 3, (4 + p^2), (6 - p).

To find:-

• p = ?

Solution:-

As we know:

• In A.P series the Common Difference is always same.

A.P :- 1, 4, 7..

So,

• D = 3 (4 - 1 = 3 and 7 - 4 = 3)

According to the question:-

→ a2 - a1 = 3 - a2

→ (4 + p^2) - 3 = (6 - p) - (4 + p^2)

→ 4 + p^2 - 3 = 6 - p - 4 - p^2

→ 1 + p^2 - 6 + p + 4 + p^2 = 0

→ 2p^2 + p - 1 = 0

→ 2p^2 + (2 - 1) p - 1 = 0

→ 2p^2 + 2p - p - 1 = 0

→ 2p (p + 1) - 1 (p + 1)=0

→ (p + 1) and (2p - 1)

→ p = -1 and p = 1/2

Hence,

• p = -1
• p = 1/2

Answer 2

Answer:

The value of P is $\frac{1}{3}$

Explanation:

• Arithmetic Progression is a sequence of numbers such that the difference between the consecutive numbers is constant.

• Sequence of numbers in the Arithmetic series could be $a, a+d, a+2d, a+3d$ etc. where,

          a = The first number in the series

          d= common difference between the numbers which is constant

• Example of a arithmetic series could be $5,11,17,23,29,35$ etc.

The given sequence of numbers is 3, 4+P, 6-P. Since the difference between the two numbers is constant, the numbers can be equated by their difference.

∴$\\(4+P-3)=(6-P-(4+P))$

  $(P+1)=(6-P-4-P)$

  $(P+1)=(2-2P)$

  $P+2P=2-1$

  $3P=1$

   $P=\frac{1}{3}$

Hence, the value of P = $\frac{1}{3}$

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