A) \[{{p}^{2}}+{{q}^{2}}=p{{'}^{2}}+q{{'}^{2}}\]
B) \[pq=p'q'\]
C) \[{{p}^{2}}-{{q}^{2}}=p{{'}^{2}}-q{{'}^{2}}\]
D) None of these
Correct Answer: C
Solution :
[c] \[2b=a+c;\] a, p, b, q, c, are in A.P. Hence \[p=\frac{a+b}{2}\] and \[q=\frac{b+c}{2}\] Again, a, p', b, q' and c = are in G.P. Hence, \[p'=\sqrt{ab}\] and \[q'=\sqrt{bc}\] \[{{p}^{2}}-{{q}^{2}}=\frac{(a-c)(a+c+2b)}{4}\] \[=\frac{(a-c)(2b+2b)}{4}\] \[[\because \,\,a+c=2b]\] \[=(a-c)\,b=ab-bc=p{{'}^{2}}-q{{'}^{2}}\]
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