An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1.
The initial term of an arithmetic progression can be called a_1 and the common difference of successive members d.
Statement 1 gives us the sum of two members of a progression, but we don't know where the progression starts, a_1, or the distance between members, d. So those two numbers could have infinite different values.
They could be 31 and 33, from a progression the d of which is 2 and the a_1 of which is 19.
They could be 30 and 34, from a progression the d of which is 4 and the a_1 of which is 6.
They could even be 32 and 32, from a progression of which the d is 0 and the a_1 is 32.
I'll show you more on how I figured out the parameters of those progressions as I go on.
Anyway, there's no way to determine what are the first ten terms or the sum of those terms, and so Statement 1 is insufficient.
Statement 2 gives us only the first term. The other terms could be pretty much any numbers. So clearly Statement 2 is insufficient.
Combined the statements give us something new.
If the first term, a_1, is 6, then a_7 = 6 + 6d and a_8 = 6 + 7d.
So adding the 7th and 8th terms, we get 6 + 6 + 6d + 7d = 64 13d = 52 d = 4.
So now we have a_1, 6, and d, 4, and we can figure out what the first ten terms of the progression are and add them up.
We don't need to do that to answer this question though. We merely need to know that we can.
So combined the statements are sufficient and the correct answer is C.
