Question

p,r,s,t,u
An arithmetic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant. If the list of letters shown above is an arithmetic sequence, which of the following must also be an arithmetic sequence?
I.2p,2r,2s,2t,2u
II.p-3,r-3,s-3,t-3,u-3
III.${p^2},{r^2},{s^2},{t^2},{u^2}$
A.I only
B.II only
C.III only
D.I and II
E.II and III

Answer 1

We can find an equation for the difference between consecutive terms from the given arithmetic sequence. Then we can take each of the given sequences and prove that the differences between two consecutive terms are equal. Then we can check whether the given sequence is arithmetic or not by the result. Then we can compare the options to get the correct answer.

Complete step-by-step answer:
It is given that p,r,s,t,u is an arithmetic sequence. We know that the difference of the term and the preceding term is constant.
So, for the given sequence, we can write,
$\Rightarrow r - p = s - r$ … (1)
Now we can consider the sequence 2p,2r,2s,2t,2u.
If this is an arithmetic sequence, the difference between the 2nd and 1st terms must be equal to the difference between 3rd and 2nd terms.
So, we need to prove that $2r - 2p = 2s - 2r$
We can take the LHS and take the common term 2 outside the bracket.
$\Rightarrow 2r - 2p = 2\left( {r - p} \right)$
On substituting value from equation 1, we get,
$\Rightarrow 2r - 2p = 2\left( {s - r} \right)$
Then we can expand the bracket,
$\Rightarrow 2r - 2p = 2s - 2r$
Therefore, sequence I is an arithmetic sequence.
Now we can consider the sequence p-3,r-3,s-3,t-3,u-3.
If this is an arithmetic sequence, the difference between the 2nd and 1st terms must be equal to the difference between 3rd and 2nd terms.
So, we need to prove that $\left( {r - 3} \right) - \left( {p - 3} \right) = \left( {s - 3} \right) - \left( {r - 3} \right)$
On expanding the LHS and RHS, we get,
$\Rightarrow r - 3 - p + 3 = s - 3 - r + 3$
On simplification, we get,
$\Rightarrow r - p = s - r$
It is the same as equation (1)
So, sequence II is also an arithmetic sequence.
Now we can consider the sequence ${p^2},{r^2},{s^2},{t^2},{u^2}$ .
If this is an arithmetic sequence, the difference between the 2nd and 1st terms must be equal to the difference between 3rd and 2nd terms
So, we need to prove that ${r^2} - {p^2} = {s^2} - {r^2}$
Let us assume ${r^2} - {p^2} = {s^2} - {r^2}$ is true.
On expansion using the identity, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ , we get,
$\Rightarrow \left( {r - p} \right)\left( {r + p} \right) = \left( {s - r} \right)\left( {s + r} \right)$
On substituting value from equation (1), we get,
$\Rightarrow \left( {r - p} \right)\left( {r + p} \right) = \left( {r - p} \right)\left( {s + r} \right)$
On cancelling the common terms, we get,
$\Rightarrow \left( {r + p} \right) = \left( {s + r} \right)$
On subtracting r on both sides, we get,
$\Rightarrow p = r$
As p and r are terms of AP, they cannot be equal. So, our assumption is false. Therefore, sequence III is not an arithmetic sequence
From the above results, only I and II are arithmetic sequences.
So, the correct answer is option D.

Note: Alternate solution to this problem is given by,
It is given that p,r,s,t,u is an arithmetic sequence. We can equate it to an arithmetic sequence of consecutive natural numbers 1,2,3,4,5
Then sequence I will become, 2p,2r,2s,2t,2u
2,4,6,8,10
We can say that it is an arithmetic sequence with difference 2.
Then sequence II will become,
-2,-1,0,1,2
This is also an arithmetic sequence with difference 1.
Consider the 3rd sequence ${p^2},{r^2},{s^2},{t^2},{u^2}$ .
On giving the substitution, we get,
1,4,9,16,25
This is not an arithmetic sequence.
From the above results, only I and II are arithmetic sequences.
So, the correct answer is option D