For the two positive numbers , if and are in a geometric progression, while , 10 and are in an arithmetic progression, then is equal to ___________.
Show answer and solution
Answer: 3
Solution
From the geometric progression condition: a·(1/18) = b², so ab = 1/18... wait, let me be careful.
a, b, 1/18 in GP means b² = a·(1/18), so a = 18b².
1/a, 10, 1/b in AP means 2·10 = 1/a + 1/b, so 1/a + 1/b = 20.
Substituting a = 18b²:
1/(18b²) + 1/b = 20
1/(18b²) + 18b/(18b²) = 20
(1 + 18b)/(18b²) = 20
1 + 18b = 360b²
360b² - 18b - 1 = 0
Using quadratic formula: b = (18 ± √(324 + 1440))/720 = (18 ± √1764)/720 = (18 ± 42)/720.
Taking positive value: b = 60/720 = 1/12.
Then a = 18·(1/12)² = 18/144 = 1/8.
Therefore: 16a + 12b = 16·(1/8) + 12·(1/12) = 2 + 1 = 3.
a, b, 1/18 in GP means b² = a·(1/18), so a = 18b².
1/a, 10, 1/b in AP means 2·10 = 1/a + 1/b, so 1/a + 1/b = 20.
Substituting a = 18b²:
1/(18b²) + 1/b = 20
1/(18b²) + 18b/(18b²) = 20
(1 + 18b)/(18b²) = 20
1 + 18b = 360b²
360b² - 18b - 1 = 0
Using quadratic formula: b = (18 ± √(324 + 1440))/720 = (18 ± √1764)/720 = (18 ± 42)/720.
Taking positive value: b = 60/720 = 1/12.
Then a = 18·(1/12)² = 18/144 = 1/8.
Therefore: 16a + 12b = 16·(1/8) + 12·(1/12) = 2 + 1 = 3.
[JEE Main 2023 · 25 Jan · Shift 2]