Problem Solving

kobel51 wrote:In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 560. What is the sum of the last 5 integers in the sequence?

A) 585
B) 580
C) 575
D) 570
E) 565

Patrick's solution is definitely the fastest.
That said, here's another (slightly longer, but still effective ) approach:

Let x = 1st number
So, x+1 = 2nd number
x+2 = 3rd number
x+3 = 4th number
.
.
.
x+8 = 9th number
x+9 = 10th number

The sum of the first 5 integers is 560
x + (x+1) + (x+2) + (x+3) + (x+4) = 560
5x + 10 = 560
5x = 550
x = 110 = 1st term (111 = 2nd term, 112 = 3rd term, etc)

So, 6th term = 115, 7th term = 116, . . . 10th term = 119,

What is the sum of the last 5 integers in the sequence?
115 + 116 + 117 + 118 + 119 = [spoiler]585 = A[/spoiler]

Cheers,
Brent

kobel51 wrote:In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 560. What is the sum of the last 5 integers in the sequence?

A) 585
B) 580
C) 575
D) 570
E) 565

Solution:

In solving this problem we must first remember that when we have 10 consecutive integers we can display them in terms of just 1 variable. Thus, we have the following:

Integer 1: x
Integer 2: x + 1
Integer 3: x + 2
Integer 4: x + 3
Integer 5: x + 4
Integer 6: x + 5
Integer 7: x + 6
Integer 8: x + 7
Integer 9: x + 8
Integer 10: x + 9

We are given that the sum of the first 5 integers is 560. This means that:

x + x+1 + x+2 + x+3 + x+4 = 560

5x + 10 = 560

5x = 550

x = 110

The sum of the last 5 integers can be expressed and simplified as:

x+5 + x+6 + x+7 + x+8 + x+9 = 5x + 35

Substituting 110 for x yields:

(5)(110) + 35 = 585

Answer: A

Alternatively, because both equations have 5x in common, we know that the difference between the sum of the first five numbers and the sum of the last five numbers is the difference between (1+2+3+4) and (5+6+7+8+9). Since 35 - 10 = 25, the sum of the last 5 is 585, which is 25 more than 560.