For the two positive numbers a,b, if a,b and 1/18 are in a geometric progression, while 1/a,10 and 1/b are in an arithmetic progression, then 16a+12b is equal to _________.

Solution

Maths sir · 5 steps

Turn the GP into a = 18b² and the AP into a + b = 20ab, then solve for b.

1. GP condition. For a, b, 1/18 in GP, the middle term squared equals the product of its neighbours: b² = a/18. So a = 18b².
2. AP condition. For 1/a, 10, 1/b in AP, twice the middle term equals the sum of the outer terms: 1/a + 1/b = 20. Clearing fractions gives a + b = 20ab.
3. Substitute and solve. Put a = 18b² into a + b = 20ab. This gives 18b² + b = 360b³. Since b > 0, divide by b to get the quadratic below.

   360b² - 18b - 1 = 0 ⇒ b = (18 ± √(324 + 1440))/720 = (18 ± 42)/720

4. Pick the positive root. The positive root is b = 60/720 = 1/12. Then a = 18 × 1/144 = 1/8.
5. So. 16a + 12b = 16 × 1/8 + 12 × 1/12 = 2 + 1 = 3.

The idea to keep. Three terms in GP means middle squared equals the product of the ends. Three terms in AP means twice the middle equals the sum of the ends. Turn each condition into one equation and substitute.

Answer from the NTA final answer key. Solution by PrepWiser.