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Given,

ap=a+(p−1)d

aq=a+(q−1)d

ar=a+(r−1)d

as=a+(s−1)d

ap,aq,ar,as are in G.P

⇒aqap=asar

consider,

aqap=araq

subtracting 1 on both sides, we get,

aqap−1=araq−1

aq−apar−aq=apaq

a+(q−1)d−[a+(p−1)d]a+(r−1)d−[a+(q−1)d]=apaq

q−rp−q=apaq............(1)

now consider,

araq=asar

subtracting 1 on both sides, we get,

araq−1=asar−1

ar−asaq−ar=araq

a+(r−1)d−[a+(s−1)d]a+(q−1)d−[a+(r−1)d]=araq

r−sq−r=araq

From (1)

r−sq−r=aqap............(2)

From (1) and (2), we have,

apaq=aqap

Therefore (p−q),(q−r),(r−s) are in G.P

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