OG 247

[umaa]

OA D

[yvichman]

They tell you that an Arith. Sequence is when each term after the 1st must be equal to the sum of the preceding # plus a constant.

They tell you P R Q T U are a set of # that meet the requirement of an Arith Sequence.

Plug in Numbers that meet that requirement for P, R, S, T, U
I chose 1 ,2, 3, 4, 5, b/c they are easy to work with and it gives us 1 as the constant

P=1 R=2 S=3 T=4 U=5 is an arith. sequence b/c if you add 1 to the preceding number it gives you the following number in the sequence
(1 + P = R, 1 +R= S)

Next step would be to plug in the #'s to the 3 choices they give you and see if they also meet the arith. sequence requirement

I) gives you 2, 4, 6, 8, 10 = sum of preceeding +2 as a constant
II) gives you -2, -1, 0, 1, 2 = sum of preceeding + 1 as a constant
III) gives your 1, 4, 9, 16, 25= sum of preceeding and constant dont form any type of pattern...

I, and II only

[rajataga]

You can just plugin numbers and try...

the sequence could be, 1, 3, 5, 7, ...


so I would be
2, 6, 10, 14, ....

Just the differene between the numbers is doubled, but it still an arithmetic sequence..

II would be
-2, 0, 2, 4,....

again an arithmetic sequence...

III would be
1, 9, 25, 49,...
Here the difference between the numbers is not constant and hence this is not an arithmetic sequence.


Hence, only I and II are.

Hence D

[sonu_thekool]

An other way to look at the problem is below :

It is given that p, r, s, t u are an arithmetic sequence.

So, think 'k' is the constant

the sequence would be :

p, p+k, p+k+k, p+k+k+k, p+k+k+k+k

p, p+k, p+2k, p+3k, p+4k

I) Multiplying by 2 would make essentially the same sequence - so, TRUE
2p, 2p+2k, 2p+4k, 2p+6k, 2p+8k

II) Subtracting 3 would also make essentially same sequence - so, TRUE
p-3, p+k-3, p+2k-3, p+3k-3, p+4k-3

III) Squaring the numbers, however, would be different.
p^2, (p+k)^2 might be way off because of the squaring. so, FALSE

The point to take is that adding or multiplying any sequence with a constant basically keeps the sequence similar but squaring the elements with a constant will produce a different sequence.

Hope this helps.