Question
Mathematics Question on Functions
For three positive integers p,q,r,xpq2=yyr=zp2r and r=pq+1 such that 3,3logyx,3logzy 7logxz are in AP with common difference 21 Then r−p−q is equal to
A
6
B
2
C
12
D
−6
Answer
2
Explanation
Solution
pq2=logxλ
qr=logyλ
p2r=logzλ
logyx=pq2qr=pqr…….(1)
logxz=p2rpq2=prq2………(2)
logzy=qrp2r=qp2………(3)
3,pq3r,q3p2,pr7q2 in A.P
pq3r−3=21
r=67pq………(4)
r=pq+1
pq=6………(5)
r=7…………(6)
q3p2=4
After solving p=2 and q=3