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If p,q,p, q, and rr are in geometric progression, which of the following is true about qr,pr,qr, pr, and pqpq?

Solution

✅ Correct Option: 2

Since pp, qq, and rr are in geometric progression, there exists a common ratio kk such that:

q=pkq = pk

r=qk=pk2r = qk = pk^2


The three products are:

qr=(pk)(pk2)=p2k3qr = (pk)(pk^2) = p^2k^3

pr=p(pk2)=p2k2pr = p(pk^2) = p^2k^2

pq=p(pk)=p2kpq = p(pk) = p^2k


For three numbers to be in geometric progression, the middle term squared must equal the product of the first and third terms.

(pr)2=(p2k2)2(pr)^2 = (p^2k^2)^2

(pr)2=p4k4(pr)^2 = p^4k^4


(pq)×(qr)=(p2k)(p2k3)(pq) \times (qr) = (p^2k)(p^2k^3)

(pq)×(qr)=p4k4(pq) \times (qr) = p^4k^4


Since (pr)2=(pq)×(qr)(pr)^2 = (pq) \times (qr), the numbers pqpq, prpr, and qrqr are in geometric progression (in that order).

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