Question

If pth, qth and rth term of an AP are in GP whose common ratio is k, then root of equation (q-r)x^2+(r-p)x+(p-q)=0 other than unity is:

Answer 1

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