n−2,4n−1,5n+2 are in A.P.

First term, a1​=n−2
Second term, a2​=4n−1
Third term, a3​=5n+2

Common difference (difference of two consecutive terms),
d=a2​−a1​=a3​−a2​
4n−1−(n−2)=5n+2−(4n−1)
4n−1−n+2=5n+2−4n+1
3n+1=n+3
2n=2

n=1

So,
a1​=n−2=1−2=−1
a2​=4n−1=4−1=3
and d=a2​−a1​=3−(−1)=4

nth term of an AP is ,
an​=a+(n−1)d
where,
a=first term
d=common difference

Therefore,
a4​=a+3d=−1+12=11
a5​=a+4d=−1+16=15