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If pth , qth, rth term of an AP are in GP , then common ratio of this GP is
1. (p-q)/(q-r)
2. (q-r)/(p-q)
3. pqr
4. none of this
2 years ago

as this is in AP 2q=p+r----eq(1)
as in GP q=p.c & r=p*c2 where  c=common difference,
putting in eq(1) 2*p.c=p+p*c2 hence c^2+1-2c=0 or ( c-1)2=0 so c=1
frm(1) p-q=q-r or (p-q)/(q-r)=1=c or(q-r)/(p-q)=1=c so both can be the answer

2 years ago
Given that.  pth ,qth,rth. Terms of an ap are in gp..therefore    a+(q-1)d.            a+(r-1)d--------------------     =  -----------------(since they are     a+(p-1)d.            a+(q-1)d.      Are in gp)Now,Let    a+(q-1)d.            a+(r-1)d--------------------     =  -----------------=  k   a+(p-1)d.              a+ (q-1)dBy theorem on equal ratiosWe geta+ qd-d -(a+rd-d)----------------------------=ka+pd-d-(a+qd-d)On simplifying we get        qd-rd------------------=k       pd-qd d(q-r)----------=kd(p-q)  Therefore    (q-r)-------------=k    (P-q)Hope this will help u

2 years ago
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