If a, b, c are in HP, then a, b, c are also in GP.

Solution

Maths sir · 4 steps

One counterexample is enough to sink an 'always' statement. Try 1, 1/2, 1/3.

1. What HP means. a, b, c are in HP when their reciprocals 1/a, 1/b, 1/c are in AP. So the middle term satisfies b = 2ac/(a+c).
2. What GP needs. For a GP, the middle term must satisfy b² = ac. The HP condition does not force this, so test a simple example.
3. Counterexample. Take 1, 1/2, 1/3. The reciprocals 1, 2, 3 are in AP, so this is an HP. But b² = 1/4 while ac = 1/3, so it is not a GP.
4. So. An HP need not be a GP, so the statement is false. Both conditions hold together only when a = b = c.

The idea to keep. For a three-term progression, compare the middle term: AP gives 2b = a + c, GP gives b² = ac, HP gives b = 2ac/(a+c). All three agree only when a = b = c.

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