If a1,a2,a3... are in A.P, then ap,aq,ar are in A.P if p, q, r are in
A.P
G.P
H.P
None of these
ap,aq,ar are in A.P ⇒ap+ar=2aq Let d be the common difference a1+(p−1)d+(a1+(r−1)d=2(a1+(q−1)d 2a1+(p+r)d−2d=2a1+2qd−2d ⇒p+r=2q ⇒p,q,r are in A.P
Let a1,a2,a3......... are in A.P such that ap, aq,ar are in both A.P and GP then aq:ap is equal to