Let a1, a2, a3,... be in AP and ap, aq, ar be in GP. Then, aq:ap is equal to
A
r−pq−p
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B
q−pr−q
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C
r−qq−p
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D
none of these
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Solution
The correct option is Dr−qq−p Given a1,a2,,a3,...... are in A.P And ap,aq,ar are in G.P Also aq=a1+(q−1)dap=a1+(p−1)dar=a1+(r−1)d, where d is the common difference of A.P Let's subtract the terms to get : ar−aq=(r−q)d and aq−ap=(q−p)d On dividing, we get ar−aqaq−ap=(r−q)(q−p) But we know that aq=ap×Rar=ap×R2=aq×R Where R is the common ratio of the G.P So putting the values we get aq×R−aqap×R−ap=(r−q)(q−p) ⇒aq(R−1)aP(R−1)=(r−q)(q−p) ⇒aqaP=(r−q)(q−p)