The third term of a geometric progression is 6. Then the product of first five…

Concept: In a geometric progression, terms equally far from a central term have reciprocal powers of the common ratio. For five consecutive GP terms with third term T, the terms can be written as T/r2, T/r, T, Tr, Tr2, so their product is T5.

This works because the ratio powers cancel in symmetric pairs, leaving only repeated factors of the middle term.

1. Let the third term be 6 and the common ratio be r.

2. The first five terms are 6/r2, 6/r, 6, 6r, and 6r2.

3. Their product is (6/r2) × (6/r) × 6 × (6r) × (6r2).

4. The powers of r cancel: r-2-1+1+2 = r0 = 1.

5. The remaining product is 6 × 6 × 6 × 6 × 6 = 65.

Cross-check: The first and fifth terms multiply to 62, and the second and fourth terms also multiply to 62; multiplying those with the middle term 6 gives 62 × 62 × 6 = 65.

Therefore, the product of the first five terms is 65.