If log5 16, log5 (3x-4), log5 (3x+97/16) are in arithmetic progression, then x…

For three numbers in arithmetic progression (A.P.), twice the middle term equals the sum of the first and third terms.

Therefore,

2log₅(3x − 4) = log₅16 + log₅[(3x + 97)/16]

Using the logarithmic property:

log₅a + log₅b = log₅(ab)

we get

2log₅(3x − 4) = log₅[16 × (3x + 97)/16]

⇒ 2log₅(3x − 4) = log₅(3x + 97)

Again using the property:

2log₅(3x − 4) = log₅[(3x − 4)²]

Hence,

log₅[(3x − 4)²] = log₅(3x + 97)

Since the bases are the same,

(3x − 4)² = 3x + 97

Expanding:

9x² − 24x + 16 = 3x + 97

9x² − 27x − 81 = 0

Divide by 9:

x² − 3x − 9 = 0

Using the quadratic formula:

x = [3 ± √45]/2

x = (3 ± 3√5)/2

Now check the domain condition:

3x − 4 > 0
⇒ x > 4/3

The root

x = (3 − 3√5)/2

is negative and hence invalid.

Thus,

x = (3 + 3√5)/2 ≈ 4.85

Since 4.85 is closest to 5, the correct option is:

✓ Option C: 5