For the sequence a1, a2, a3 ... an, an is defined by an=n+2 for all

Question

For the sequence a1, a2, a3 ... an, an is defined by \(a_n=n+2\) for all even integers and \(a_n=n−1\) for all odd integers for all integers n≥1. What is the sum of the first 10 terms of this sequence?

A. 20
B. 40
C. 56
D. 60
E. 62

A. 20
B. 40
C. 56
D. 60
E. 62

GMATinsight · Accepted Answer

As per the formulas quoted by the questions

a_1 = 1-1 = 0  
  a_2 = 2+2 = 4  
  a_3 = 3-1 = 2  
  a_4 = 4+2 = 6  
  a_5 = 5-1 = 4  
  a_6 = 6+2 = 8  
  a_7 = 7-1 = 6  
  a_8 = 8+2 = 10  
  a_9 = 9-1 = 8  
  a_{10} = 10+2 = 12

Sum m =  (0+2+4+6+8) + (4+6+8+10+12)  =  20  +  40  = 60

Answer: option D

SVP482 · Answer

First 10 digits contain 5 odd and 5 event. The odd start with n - 1 and even starts with n + 2

odd - 0, 2, 4, 6 and 8 ( 5 values) 
Event - 2, 4, 6, 8, 10

Adding them 10 + 2(20) = 60.

ScottTargetTestPrep · Answer

We are given the following sequence:

a(n) = n + 2 for all even integers n, and a(n) = n - 1 for all odd integers n. We need to determine the sum of the first 10 terms of the sequence. Let’s start with a(1).

a(1) = 0

a(2) = 4

a(3) = 2

a(4) = 6

a(5) = 4

a(6) = 8

a(7) = 6

a(8) = 10

a(9) = 8

a(10) = 12

Thus, the sum is 60.

Alternate solution:

If we assume that a(n) = n, regardless of whether n is odd or even, then the sum of the first 10 terms will be the sum of the first 10 positive integers. That is, the sum will be 1 + 2 + 3 + … + 10 = 55. Notice that there are 5 even and 5 odd numbers in the first 10 positive integers. However, since we are given that a(n) = n + 2 for even values of n and a(n) = n - 1 for odd values of n, for each even value of n we need to add 2 and for each odd value of n we need to subtract 1. That is, we need to modify the sum (55) by adding 5 x 2 = 10 to it and subtracting 5 x 1 = 5 from it. Thus, the sum we seek is:

55 + 10 - 5 = 60

Answer: D

bumpbot · Answer

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For the sequence a1, a2, a3 ... an, an is defined by an=n+2 for all

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