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

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