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Correct option is B)

p, q and r are in A.P

i.e. $2q=p+r$

General term of G.P is $T(K)=ar_{(kβ1)}$

$p_{th}$ term is :- $T(p)=ar_{(pβ1)}...(1)$

$q_{th}$ term is :- $T(q)=ar_{(qβ1)}...(2)$

$r_{th}$ term is :- $T(r)=ar_{(Rβ1)}...(3)$

To prove that ;- $T(p),T(q),T(r)$ are in G.P

$T(q)_{2}=T(p).T(r)$

$[ar_{(qβ1)}]_{2}=ar_{(pβ1)}.ar_{(Rβ1)}$

$a_{2}ar_{2(qβ1)}=a_{2}r_{(pβ1)}r_{(Rβ1)}$

$r_{2(qβ1)}=r_{(P+Rβ2)}$

Taking only powers

$2(qβ1)=P+Rβ2$

$2qβ2=P+Rβ2$

$2q=P+R$

This proves that $p_{th},q_{th}$ and $r_{th}$

term are in G.P

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