If log10 2, log10 (2x- 1) and log10(2x+ 3 ) are three consecutive terms of an A.P, then the value of x isa)1b)log52c)log2 5d)log105Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Solution:

Given, log10 2, log10 (2x- 1) and log10(2x + 3) are three consecutive terms of an A.P.

Let us assume that the common difference of this A.P. be d.

Then, we know that for any three consecutive terms of an A.P., the middle term is the arithmetic mean of the first and the third terms.

So, we have:

log10 (2x- 1) = (log10 2 + log10 (2x + 3))/2

log10 (2x- 1) = log10 √(2(2x + 3))

Taking antilogarithms on both sides, we get:

2x - 1 = √(2(2x + 3))

Squaring both sides, we get:

4x² - 8x - 5 = 0

On solving this quadratic equation, we get:

x = (8 ± √(64 + 80))/8

x = (8 ± 4√5)/8

x = 1 ± √5/2

Since x cannot be negative, we have:

x = 1 + √5/2

x = (1/2)log2 5

Hence, the correct option is (c) log2 5.

Answer

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