This question asks us to find the sum of a series with the first item: a1=101, last item: an=998and tolerance: d=3

first, we need to find n:

an=a1+d*(n-1)

998=101+3*(n-1) =>n=300

Sn=(a1+an)*n/2=(101+998)*300/2=164,850

billzhao wrote:This question asks us to find the sum of a series with the first item: a1=101, last item: an=998and tolerance: d=3

first, we need to find n:

an=a1+d*(n-1)

998=101+3*(n-1) =>n=300

Sn=(a1+an)*n/2=(101+998)*300/2=164,850

agree with you..


nice to recall sum of series formula..

Sum of AP=n/2(2a+(n-1)d
Sum of GP=a(1-r^n)/1-r *

Billzhao,

Your soln seems neat and clean, but how did you conclude that
a1=101 and the rest. The formula looks straightforward. If you could please elaborate it would be great. This is how tried to solve:

let a, b, c be the 3 numbers, so a+b+c=?
now, abc-3x=2
=> abc = 2+3x

Comparing the values in A, B, C, D, E simulatenously w/-
30, 27, 24, 21, 18, 15, 12, 9, 3 (I tried to add 2 to these values and see the units digit matches to either options and if it does is it divisible by 3)

If not then check the remainder. By trial and error and canceling the options that did not fit the criteria I reached the same answer as yours.

But an explanation would help me understand ur soln.

kanha81,

The Cliffnotes Guide for the GMAT (Chapter 15: Sequences and Series) is a pretty good section that lays out the formulas being used for solving such questions. Their questions are not GMAT level but the formulas and explanations provide a foundation that makes it easier to solve these types of questions. You can get this book from any library or bookstore. Hope that helps.