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If the list of letters $P$, $R$, $S$, $T$, $U$ is an arithmetic sequence, which of the following are also in arithmetic sequence?

1. $2P, 2R, 2S, 2T, 2U$
2. $P-3, R-3, S-3, T-3, U-3$
3. $P^2, R^2, S^2, T^2, U^2$
1. I only
2. I and II
3. II and III
4. I and III

### 1 comment

Before checking answer take -> P =  a - 2*d, Q  =  a - d, R = a  ,  S = a + d and verify options.

Answer is $B$ because If we multiply the terms of an $AP$ with a constant, the common difference will get multiplied by the constant  and if we subtract the terms of an $AP$ with a constant, the common difference will be same but if we make square of $AP$ element  then difference will not be same.

if we multiply by any no to AP than difference between them will b same

No. If we multiply the terms of an AP with a constant, the common difference will get multiplied by the constant too.

Good Explanation!

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

1. $2P,2R,2S,2T,2U$ are $2,4,6,8,10\implies \text{AP}$
2. $P-3,R-3,S-3,T-3,U-3$  are $-2,-1,0,1,2\implies \text{AP}$
3. $P^{2},R^{2},S^{2},T^{2},U^{2}$  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 $(B).$

### 1 comment

Nice Explanation with example.

by taking the arithmetic mean of each option , if it is AP then arithmetic mean property must satisfied

a+c= 2b.

by

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.

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

$ans:B$

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