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concepts of sets

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concepts of sets

by romitvsingh » Sat Nov 26, 2011 12:21 am
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 be an 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

A) 1 only
B) 2 only
C) 3 only
D) 1 & 2
E) 2 & 3

pls explain in detail
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by shankar.ashwin » Sat Nov 26, 2011 12:42 am
Frame your own series to satisfy the conditions; Let the first term be '1' and the constant be '1'

So series is 1,2,3,4,5 (p,q,r,s,t)

1) 2,4,6,8,10 - An A.sequence (constant is '2')

2) -2,-1,0,1,2 - An A.sequence (constant is '1')

3) 1,4,9,16,25 - Not an Arithmetic sequence.

So, (1) and (2) are true. D
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by tpr-becky » Sat Nov 26, 2011 2:24 pm
You can certainly use Pluggin in (or picking your own numbers) to try this type of problem, and during the exam I think this would be the best approach - see above.

However, in studying it is a good idea to understand the concept being tested.

When you have a set of numbers in which each number is the number before plus a constant - so you can rewrite it:
p, p+k, p+k+k, etcc

when you multiply each of the numbers in the set by 2 then the distributive law shows that you multiply each term by 2 - thus it will still be the number before plus a constant (that constant will just be 2 times the original constant.
When you subtract three from each of the terms, you will get a number plus a constant -3 (thus it will still be the previous number plus a constant).

however, when you square a term you have to add it first - you won't simply be adding the squared constant - for instance (2+1)^2 is not the same as 2^2 + 1^2 thus the constant will not remain the same across the set.

A good question for the distributive property
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