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15 votes

Best answer

7 votes

Lets take $P,R,S,T,U$ are $1,2,3,4,5\implies \text{AP}$

- $2P,2R,2S,2T,2U$ are $2,4,6,8,10\implies \text{AP}$
- $P-3,R-3,S-3,T-3,U-3$ are $-2,-1,0,1,2\implies \text{AP}$
- $P^{2},R^{2},S^{2},T^{2},U^{2}$ are $1,4,9,16,25\implies {\color{Red} {\text{Not an AP}}}$

$\therefore (I)\: \text{and}\: (II)$ are in $\text{AP}.$

So, the correct answer is $(B).$

0 votes

If all the terms of an AP be multiplied or divided by the same quantity, the resulting terms will form an AP, but with a new common difference, which will be the multiplication/division of the old common difference.

$I: 2P,2R,2S,2T,2U$

If the same quantity be added to, or subtracted from all the terms of an AP, the resulting terms will form an AP, but with the same common difference as before.

$II: P-3,R-3,S-3,T-3,U-3$

$ans:B$