For each positive integer n, let an = n(-1)^n(x-1). If the sum of

Question

For each positive integer n, let  a_n = n(-1)^n(x-1) . If the sum of  a_1  through  a_{20}  is equal to 60, what is the value of x?

(A) -5
(B) -2
(C) 2
(D) 4
(E) 7

gmatophobia · Accepted Answer

a_1 = 1(-1)^1(x-1) = -1(x-1)

a_2 = 2(-1)^2(x-1) = 2(x-1)

a_3 = 3(-1)^3(x-1) = -3(x-1)

a_4 = 4(-1)^4(x-1) = 4(x-1)

.
.
General Pattern

a_n = -n(x-1)  ⇒ n = odd
 
  a_n = n(x-1)  ⇒ n = even

.
.

a_{19} = -19(x-1)

a_{20} = 20(x-1)

a_1 + a_2 + a_3 + ...+ a_{19}+a_{20} = -1(x-1) + 2 (x-1) -3 (x-1) + 4 (x-1) -5 ....... -19(x-1) + 20(x-1) = 60

(x-1)(-1 + 2 -3 + 4 -5 + 6 ... -19 + 20) = 60

(x-1)(1 + 1 + 1 + 1 + 	ext{upto 10 times}) = 60

(x-1)(10) = 60

(x-1) = 6

x = 7

Option E

Samé · Answer

Bunuel 
Please, where am I wrong ?

the sum of a1 through a20 is equal to 60 means the average of the set is 60/20 = 3
the average is also a10 + a11 = 3*2=6
Then 10(x-1)-11(x-1)=6 --> x=-5

Bunuel · Answer

The red part above is not correct. The average of a sequence does not always equate to the average of the two middle terms. This is true only for specific sequences, such as arithmetic progressions. However, the given sequence is not of that kind. The sequence is:

-6, 12, -18, 24, -30, 36, -42, 48, -54,  60, -66,  72, -78, 84, -90, 96, ...

muskaaaanmehra · Answer

a1 = -1(x-1), a2 = 2(x-1), a3=-3(x-1), a4= 4(x-1).........a20=20(x-1)

Sum of first 20 elements = (x-1)

Sum of even elements = 10*(2+20)/2=110
Sum of odd elements = 10*(19+1)/2 = 100

Which finally gives us 10(x-1) = 60

x=7

musaraza175 · Answer

Can someone guide me, why can't I solve this through the sum formula i.e (N/2*(a1+a20))=60

Bunuel · Answer

That formula for the sum applies when the sequence of numbers is in arithmetic progression. However, if you carefully review the discussion above, you’ll see that the sequence in question is not an arithmetic progression, so the formula cannot be applied.

Hope this helps.

adityaprateek15 · Answer

For n=1, an = -1(x-1) = 1-x
For n=2, an = 2(x-1)
 
a2-a1 = 2(x-1) - 1(x-1) = x-1

This pattern repeats for every 2 consecutive values of an and there are 10 such patterns from a1 to a20

So, 10(x-1)=60, x-1 = 6 => x=7

D

For each positive integer n, let an = n(-1)^n(x-1). If the sum of

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GMAT Problem Solving (PS) Questions

For each positive integer n, let an = n(-1)^n(x-1). If the sum of

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