Let the common difference of the A.P. be d and the common ratio of the G.P. be r.
Given a1=1 and g1=a1=1.
From a2+g2=1, we have:
(a1+d)+g1r=1
1+d+r=1⇒d=−r
From a3+g3=4, we have:
(a1+2d)+g1r2=4
1+2d+r2=4
2d+r2=3
Substituting d=−r into the equation:
−2r+r2=3
r2−2r−3=0
(r−3)(r+1)=0
Since the G.P. is increasing and g1=1, the common ratio r must be greater than 1. Thus, r=3.
Then, d=−3.
We need to find a10+g5:
a10=a1+9d=1+9(−3)=−26
g5=g1r4=1(3)4=81
a10+g5=−26+81=55
Answer: 55