(d) : Since, $2, p$ and $q$ are in G.P.\n\n\nLet first term of the A.P. be $a$ and common difference be $d$.\n\nand $T_{13}=a+12 d=q$\n∴ From (ii) and (iii), we get $d=p-2$\nFrom (ii) and (iv), we get $6 d=q-2 \Rightarrow 6(p-2)=q-2$\n\n\nFrom (i) and (v), we get $p=2$ or $p=10 \therefore q=2$ or 50 But $q \neq 2$ Hence, $p=10, q=50 \therefore d=8$ and $a=-46$\n\nSince, $5^{\text {th }}$ term of G.P. $=n^{\text {th }}$ term of A.P.\n