If 2p−1,3p+1,11 are terms in AP, thena2​−a1​=a3​−a2​ .........(1)From given:a1​=2p−1a2​=3p+1a3​=11From (1), we get(3p+1)−(2p−1)=11−(3p+1)3p+1−2p+1=11−3p−1p+2=10−3p4p=8p=2For p=2, these terms are in AP.And the numbers are (2⋅2−1),(3⋅2+1),11i.e. 3,7,11