Let 3, a, b, c be in A.P. and 3, a-1, b+1, c+9 be in G.P. Then, the arithmetic mean of a, b and c is :

Solution

Maths sir · 4 steps

Write everything in terms of the common difference d, then let the G.P. condition fix d.

1. One unknown. With common difference d: a = 3 + d, b = 3 + 2d, c = 3 + 3d. The G.P. then reads 3, 2 + d, 4 + 2d, 12 + 3d.
2. Middle-term rule. In a G.P. the square of a term equals the product of its neighbours. Use the first three terms.

   (2+d)² = 3(4+2d) ⇒ d² - 2d - 8 = 0 ⇒ d = 4 or d = -2

3. Reject d = -2. With d = -2 the sequence is 3, 0, 0, 6. A G.P. cannot have a zero term, and the ratio jumps from 0 to 6, so d = 4. Check: 3, 6, 12, 24 is a G.P. with ratio 2.
4. So. a = 7, b = 11, c = 15, and their mean is (7 + 11 + 15)/3 = 33/3 = 11. The answer is (d).

The idea to keep. When a quadratic gives two values, put each one back into the original sequence. A G.P. can never contain a zero term, so any root that makes a term zero must be rejected.

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