Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r², then r² - d is equal to :

Solution

Maths sir · 4 steps

Write the GP as a, ar, ar², turn the doubled middle into an AP condition, and r falls out first.

1. AP condition. After doubling, the numbers are a, 2ar, ar². In an AP the middle term is the average of the outer two, so 2(2ar) = a + ar².

   r² - 4r + 1 = 0 ⇒ r = 2 ± √3

2. Pick the root. The fourth term condition ar³ = 3r² gives a = 3/r > 0. With a positive first term the GP increases only when r > 1, so r = 2 + √3, 1/r = 2 - √3 and r² = 7 + 4√3.
3. Find d. The AP is 3/r, 6, 3r. So d = 6 - 3/r = 6 - 3(2 - √3) = 3√3.
4. So. r² - d = 7 + 4√3 - 3√3 = 7 + √3, which is option (b).

The idea to keep. For three terms of a GP or AP, write them as a, ar, ar² and use the middle-term rule. The word "increasing" is there to choose between the two roots, so always use it.

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