Problem Solving

If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

517
518
519
521
525

Please show your thought process. Thank you!

18 consecutive odd integers:

x, x+2...x+2n--> x, x+2...x+34

Sum of an arithmetic sequence with the first term being 1 and the last term being 34.

n= total number of terms in the sequence.

Sn= 17/2(36)=17*18=306

18x+306/18=534

18x+306=9612

x=517 A

Since the sequence is evenly spaced, we know that the mean=median=mean of first and last term.

Let x = first term.
Since there are 18 terms in the sequence, and the sequence is made up of odd consecutive terms, we know that the last term can be represented as x + 2*17 = x + 34.

mean = mean of first and last term gives us:
534 = (x + x +34)/2
534 = x + 17
x = 534 - 17
x = 517

Another approach is that since there are 18 terms, the median is equal to the average of the two middle terms (terms 9 and 10), denote these two terms T9, T10.

534 = mean = median = T9+T10/2.

Since T9 and T10 are consecutive integers, T9 = 533 and T10 = 535.

Since the terms are consecutive, odd integers, keep subtracting 2 to get back to T1.

T9 = 533
T8 = 531
T7 = 529
T6 = 527
T5 = 525
T4 = 523
T3 = 521
T2 = 519
T1 = 517

This second approach might seem longer, but it may be more intuitive. Using your first approach, you must make sure that you figure out what the last term is in terms of the first term (initial thought might be x + 36).

If you aren't comfortable making these moves with evenly spaced number sets, then check out Manhattan GMAT's Number Properties book.

sum of 18 odd nos=534*18

Sn=(n/2)(2a+(n-1)d)
here n=18
d=2, a =least number
(18/2)(2a+(18-1)2)=534*18
2a+34=1068
a=517
A

This is easy ,

The answer should be odd so it cannot be B

consecutive odd integers are in a AP

and the average is 534 so we know the 9th value ( =533)

Hence the smallest or the first value ( in ascending order ) is

first + 16 = 533 ; first = 517