In an increasing sequence of 8 consecutive odd integers

Hi,

Let the first number be a...
Then the Sum of first 4 numbers are a+a+2+a+4+a+6=4a+12=656...
Sum of last four numbers= a+8+a+10+a+12+a+14=4a+44=(4a+12)+32=656+32=688..
A

A different approach:

Let's examine 8 consecutive odd integers: 5, 7, 9, 11, 13, 15, 17, 19

Notice that 13 is 8 greater than 5
And 15 is 8 greater than 7
And 17 is 8 greater than 9
And 19 is 8 greater than 11

So, each each value in the LAST 4 integers is 8 greater than its corresponding integer in the FIRST 4 integers.

So, sum of LAST 4 integers = sum of FIRST 4 integers + 8 + 8 + 8 + 8
= 656 + 8 + 8 + 8 + 8
= 656 +32
= 688

Answer:

GMATPrepNow or chetan2u or Bunuel or anyone, I have a question.

When writing the terms for an equation containing consecutive odd integers . . .

. . . is there any rule about when to use (x, x + 2, x + 4 ...) versus when to use (2n + 1, 2n + 3, 2n + 5 ...)?

Posters here have used both.

I have not seen this issue addressed very clearly; the math gurus I've found all say "use either," without explaining why on earth you'd take the longer route (2n + 1).

I understand that 2n + 1 guarantees an odd integer. That said:

1) I have yet to see a problem where the simpler form (x, x+2...) did not work, but maybe that's just dumb luck; and

2) why would the few people I have found not just say, "You can use either, but the 'x, x + 2' approach typically involves less calculation"?

The best strategy for me:

median = avg. for consecutive integers,
the sum is average x terms.

656/4 = 164 so that means the first 4 terms are 161, 163, 165, 167

the last 4 terms are 169, 171, 173, 175 with the median/avg. being 172
172x4 = 688

answer A

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