sequence - 3

[guerrero]

For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series?

A. 5,250
B. 10,098
C. 14,850
D. 15,147
E. 15,150

OA D

[GMATGuruNY]

For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series?

A. 5,250
B. 10,098
C. 14,850
D. 15,147
E. 15,150

OA D

WRITE IT OUT until you see the PATTERN.

First 4 terms of the sequence:
1+2+3
2+3+4
3+4+5
4+5+6

Last 4 terms of the sequence:
96+97+98
97+98+99
98+99+100
99+100+101

The first value (1) and the last value (101) each appear only once.
1+101 = 102.

The second value (2) and the penultimate value (100) each appear two times.
2+2+100+100 = 204.

Every integer between 3 and 99, inclusive, appears THREE TIMES.

For any set of consecutive integers:
The number of integers = biggest - smallest + 1.
The average of the integers = (biggest + smallest)/2.
The sum of the integers = number * average.

Thus, for the consecutive integers between 3 and 99, inclusive:
Number = 99-3+1 = 97.
Average = (99+3)/2 = 51.
Sum = 97*51 = 4947.
Since each of these integers appears three times, we multiply by 3:
3 * 4847 = 14,841.

Thus:
Total sum = 102 + 204 + 14841 = 15,147.

The correct answer is D.

[Anju@Gurome]

For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series?

In general, k-th term = k + (k + 1) + (k + 2) = 3k + 3

Hence, the required sum = (3*1 + 3) + (3*2 + 3) + (3*3 + 3) + ... + (3*99 + 3) = (3*1 + 3*2 + 3*3 + ... 3*99) + (3 + 3 + 3 ... 99 times) = 3*(1 + 2 + 3 + ... + 99) + 3*99

Now, (1 + 2 + 3 + ... + 99) = 99*100/2 = 99*50

So, the required sum = 50*99 + 3*99 = 53*99 = Some integer with units digit 7

The correct answer is D.

[vipulgoyal]

1+2+3 + 2+3+4 + 3+4+5 = .....99+100+101
6 + 9 + 12 + 15....300
a+(n-1)d
6+(n-1)d= 300
n= 99
S=99/2(6+300)
=15147

[CSASHISHPANDAY]

The nth term of the sequence is the sum of three consecutive integers starting with n
=> T(n) = n + (n+1) + (n+2)
=> T(n) = 3(n+1)

Therefore S(n) = 3*[(n(n+1)/2 + n] = (3/2)*[n(n+3)]
=> S(99) = (3/2)*[99*102] = 3*99*51 = 15,147

Option (D)

[Scott@TargetTestPrep]

For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series?

A. 5,250
B. 10,098
C. 14,850
D. 15,147
E. 15,150

OA D

Solution:

Let’s write out the first couple of terms of this sequence:

1st term = 1 + 2 + 3 = 6

2nd term = 2 + 3 + 4 = 9

3rd term = 3 + 4 + 5 = 12

and so on.

We can see that the 1st term is 3 x 2 = 6, 2nd term is 3 x 3 = 9, 3rd term is 3 x 4 = 12. Thus the nth term is 3 x (n + 1), and the 99th term is therefore 3 x 100 = 300. We need to find the sum of the following:

6 + 9 + 12 + … + 300

We see that is a sum of an arithmetic sequence, which we can use the following formula:

Sum = average x quantity

Sum = (6 + 300)/2 x 99

Sum = 153 x 99

Sum = 15,147

Alternate Solution:

Let’s write out the first couple of terms of this sequence:

1st term = 1 + 2 + 3 = 6

2nd term = 2 + 3 + 4 = 9

3rd term = 3 + 4 + 5 = 12

…

99th term = 99 + 100 + 101

The sum of the first column is 1 + 2 + 3 + … + 99 = (99 x 100)/2 = 99 x 50 = 4,950. Each term in the second column is one greater than the corresponding therm in the first column; thus the sum of the second column is 4,950 + 99 = 5,049. Similarly, each term in the third column is one greater than the corresponding term in the second column. Thus, the sum of the third column is 5,049 + 99 = 5,148. Adding the sums together, we find that the sum of all terms is 4,950 + 5,049 + 5,148 = 15,147.

Answer: D