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16 Sep 2021, 01:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
66% (01:04) correct
34%
(01:29)
wrong
based on 105
sessions
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Not Attempted Yet
\(a, \ b, \ c, \ d, \ e\)
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 numbers shown above is an arithmetic sequence, which of the following must also be an arithmetic sequence?
I. \(a − 3, \ b − 3, \ c − 3, d − 3, \ e − 3\)
II. \(a^3, \ b^3, \ c^3, \ d^3, \ e^3\)
III. \(4a, \ 4b, \ 4c, \ 4d, \ 4e\)
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
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 numbers shown above is an arithmetic sequence, which of the following must also be an arithmetic sequence?
I. \(a − 3, \ b − 3, \ c − 3, d − 3, \ e − 3\)
II. \(a^3, \ b^3, \ c^3, \ d^3, \ e^3\)
III. \(4a, \ 4b, \ 4c, \ 4d, \ 4e\)
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
16 Sep 2021, 01:43
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Bunuel
Solution:
We are told that \(a, \ b, \ c, \ d, \ e\) are in AP. This means we can say \(b-a=c-b=d-c=e-d=k\) (some constant).
Now, let us pick each series one and one and see if they too follow the trend of "a term is equal to the sum of the preceding term and a constant."
I: \(a − 3, \ b − 3, \ c − 3, d − 3, \ e − 3\)
let us take the first 2 of them and subtract: \(b-3-(a-3)\)
\(=b-3-a+3\)
\(=b-a\) which we know \(=k\) (some constant).
A similar trend will be followed for the rest of them. Thus \(a − 3, \ b − 3, \ c − 3, d − 3, \ e − 3\) is an AP.
We can eliminate options B and C.
II: \(a^3, \ b^3, \ c^3, \ d^3, \ e^3\)
This can get complicated because if we do \(b^3-a^3\), we must know the identity \(x^3-y^3=(x-y)(x^2+xy+y^2)\). BUt let us suppose we don't know this identity.
Another way to go about this is to assume values.
Let us assume \(a, \ b, \ c, \ d, \ e=1,2,3,4,5\) (in AP with \(k=1\)) respectively. So, \(a^3, \ b^3, \ c^3, \ d^3, \ e^3=1,8,27,64,125\) respectively.
Now we can very easily see that 1,8,27,64,125 are not in AP because \(8-1≠27-8≠64-27≠125-64\). We can eliminate option E.
III: \(4a, \ 4b, \ 4c, \ 4d, \ 4e\)
let us take the first 2 of them and subtract: \(4b-4a\)
\(=4(b-a)\)
\(=4k\) which we know is some constant.
A similar trend will be followed for the rest of them. Thus \(4a, \ 4b, \ 4c, \ 4d, \ 4e\) is an AP.
Hence the right answer is Option E.
28 Aug 2025, 20:25
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