CAT 2003 Slot 1QA Question 40

Arithmetic ProgressionEasy

If log3 2, log3 (2x − 5), log3 (2x − 7/2) are in arithmetic progression, then the value of x is equal to

Answer & solution

Correct answer: 3

  • A

    5

  • B

    4

  • C

    2

  • 3

Solution

log3 2, log3 (2x − 5), log3 (2x − 7/2) are in A.P.

∴ 2 × log3 (2x − 5) = log3 2 + log3 (2x − 7/2)

∴ log3 (2x − 5)2 = log3 [2 × (2x − 7/2)]

Let 2x = a, then we have,

(a − 5)2 = 2 × (a − 7/2)

a2 − 10a + 25 = 2a − 7

a2 − 12a + 32 = 0

a2 − 8a − 4a + 32 = 0

(a − 8)(a − 4) = 0

a = 8 or 4

2x = 8 or 2x = 4

x = 3 and x = 2

x = 2 cannot be the answer as (2x − 5) would become negative and logarithms of negative numbers are not defined.

∴ x = 3

Hence, option (d).

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CAT 2003 Slot 1 QA Q40: If log 3 2, log 3 (2 x − 5), log 3 (2 x − 7/2) are in arithmetic progression, then the value of x — Solution | TheCATExam