Let three real numbers a, b, c be in arithmetic progression and a+1, b, c+3 be in geometric progression. If a>10 and the arithmetic mean of a, b and c is 8, then the cube of the geometric mean of a, b and c is

Solution

Maths sir · 4 steps

Mean 8 fixes b = 8; write a, c as 8 - d, 8 + d and let the GP condition pick d.

1. Use the AM. The mean of a, b, c is 8, so a + b + c = 24. In an AP, a + c = 2b, so 3b = 24 and b = 8. Write a = 8 - d and c = 8 + d.
2. Use the GP. For a+1, b, c+3 in GP, the middle term squared equals the product of the outer two. Here a + 1 = 9 - d and c + 3 = 11 + d.

   64 = (9-d)(11+d) ⇒ d² + 2d - 35 = 0 ⇒ (d+7)(d-5) = 0

3. Choose d. So d = 5 or d = -7. With d = 5, a = 3, which breaks a > 10. With d = -7, a = 15, b = 8, c = 1.
4. So. The geometric mean of a, b, c is ∛abc, so its cube is abc = 15 × 8 × 1 = 120, option (a).

The idea to keep. When three terms in AP have a known sum, write them as b - d, b, b + d so only d is unknown. And the cube of the GM of three numbers is just their product.

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