I solved it using patterns -
5 cities
4 , 3, 2, 1 for each city per say.
extending to 30 cities, would mean 1,2,3,4, ..............,29
thats the series of consecutive integers - now calculating the sum -
no of terms = 29-1 +1 29
Average = 1+29 / 2 = 15
sum = 29 * 15 = 435
choice B
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Hi rishi02,

Your set-up is correct, but you should double-check your math (485 is NOT the result of either of those calculations).

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we can solve it in 2 approaches:
1) find the sum of consecutive integers: (1+29/2)*29=435
or
2) how many matches are for 30 items: 2C30=29*30/2=435

1+2+3+4+5+6+7+8+9+10=55
11+12+...........20 =100+55
21+22+...........30 = 200+55
Total 465

Total - 30= 435

dave13 - I did miss the tag.
Or my email has gone berserk. Sorry.

I'm glad you're happy.
You should be! You are very close!
(Plus, I do not think this formula
is intuitive at all. I use the "easier" version. )
I uploaded the table, see below.

Notice . . .
how many dots are under City A? Zero (0)
Because we begin at City A, let's say City A = City 1
City 1? 0 dots
How many dots under City B = City 2? One (1) dot
City C = 3? 2 dots

There is one fewer dot than there are cities.
There is no distance between City A and City A
Easy mistake.

You have to account for that disparity
when you think about the first term, \(a\), in your sum
A-to-A = 0 (as a first term in your sum of integers)

So your formula will work perfectly if
you alter one of two things.

\(\frac{n}{2} (2a+(n-1)d)\)

Either start with \(a\) equal to 0, i.e.
\(a = 0\), then
\(n = 30\)
\(d = 1\)

OR let \(a = 1\), in which case
\(n = 29\) (one fewer dot/distance between than number of cities)
\(d = 1\)

Try your approach with one of these two changes --
whichever seems most intuitive to you.
Does this explanation make sense?

(Alternatively, although I think it is harder,
you could use your original answer and correct for over-counting.
You counted 30 city-city distances too many.
You have to subtract for A-A, B-B, C-C, etc.
So: (465 - 30) = 435.)

Approach - the question says - "If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?"
Essentially it is asking for pairs to be represented by 1 dot, so we can use combinations method
30C2 --> 435