Let 1/16, a and b be in G.P. and 1/a, 1/b, 6 be in A.P., where a, b > 0. Then 72(a + b) is equal to ___________.

Solution

Maths sir · 4 steps

Use the G.P. to write b in terms of a, then the A.P. gives one quadratic in a.

1. G.P. condition. For 1/16, a, b in G.P., the middle term squared equals the product of its neighbours: a² = b/16, so b = 16a².
2. A.P. condition. For 1/a, 1/b, 6 in A.P., twice the middle term equals the sum of the outer two: 2/b = 1/a + 6. Put b = 16a², so 2/b = 1/8a², and multiply through by 8a².

   48a² + 8a - 1 = 0 ⇒ a = (-8 ± 16)/96

3. Keep the positive root. Since a > 0, take a = 8/96 = 1/12. Then b = 16 × 1/144 = 1/9. Check: 12, 9, 6 is an A.P.
4. So. a + b = 3/36 + 4/36 = 7/36, and 72(a + b) = 72 × 7/36 = 14.

The common slip: 54. Taking the negative root a = -1/4 gives b = 1 and 72(a + b) = 54, but the question needs a > 0.

The idea to keep. When terms sit in a G.P. and their reciprocals in an A.P., use the G.P. to cut the unknowns down to one. Then let the sign condition pick the right root.

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