Let a1, a2, a3, be an A.P. and g1=a1, g2, g3, be an increasing G.P. If a1=a2+g2=1 and a3+g3=4, then a10+g5 is equal to:
- 81
- 76
- 62
- 55
Answer
(4) 55
Solution
a1 = g1 = 1. Let the A.P. have common difference d and the G.P. common ratio q.
a2 + g2 = 1 (1 + d) + q = 1 d = -q
a3 + g3 = 4 (1 + 2d) + q2 = 4
q2 - 2q - 3 = 0 (q - 3)(q + 1) = 0
The G.P. is increasing, q = 3 (not -1), so d = -3.
Then a10 = 1 + 9(-3) = -26 and g5 = 34 = 81, hence
a10 + g5 = -26 + 81 = 55
Answer: (4)