Discussion

Question:

The numbers −16, 2^x+3− 2^2x−1− 16, 2^2x−1+ 16 are in an arithmetic progression. Then x equals ________.

Explanation:

−16, 2^x+3− 2^2x−1− 16, 2^2x−1+ 16 are in AP

If a, b and c are in AP ⇒ 2b = a + c

∴ 2 × (2^x+3− 2^2x−1− 16) = -16 + 2^2x−1+ 16

⇒ 2^x+4− 2^2x− 32) = 2^2x−1

⇒ 16 × 2^x− 2^2x− 32 = 2^2x/2

⇒ 32 × 2^x− 2 × 2^2x− 64 = 2^2x

Take 2^x = a and 2^2x = a²

⇒ 32a− 2a²− 64 = a²

⇒ 3a² - 32a + 64 = 0

⇒ 3a² - 8a - 24a + 64 = 0

⇒ (3a - 8)(a - 8) = 0

⇒ a = 8/3 or 8

⇒ 2^x = 8/3 or 8

[8/3 is rejected as 2^x cannot be a fraction]

⇒ 2^x = 8 = 2³

⇒ x = 3

Hence, 3.

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