For every integer K from 1 to 10, inclusive the kth term of a certain sequence is given by (-1)^(k+1) (1/2^K).
If T is the sum of first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to 1/2
E. Less than ¼

Calculate until you see the pattern.
Some test-takers might find it helpful to visualize the sum on a number line.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (1/16).
In other words, the sum will alternate between increasing a little and then decreasing a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Here are the first four terms, plotted on a number line:



Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.

1<=K<=10
T1=(-1)^2*1/2=1/2
T2=(-1)^3*1/2^2=-1/2^2
T1+T2+T3+...................+T10
[{(1/2)-(1/2^2)}+{(1/2^3)-(1/2^4)}+....................+{(1/2^9)-(1/2^10)}]
= (1/2^2)+(1/2^4)+(1/2^6)+(1/2^8)+(1/2^10)
USING GP FORMULA
= (1/2^2)(1-(1/2^2)^5)/(1-(1/2^2))
=(1/4)*(2^10-1)*4/(2^10*3)
=(2^10-1)/(2^10*3)
=(1/3)*(1-(1/2^10))
=~ 1/3 = .333 (1/1024 Is negligible w.r.t two decimal place)
=.33
= between .25 and .5

sanjoy18 wrote:1<=K<=10
T1=(-1)^2*1/2=1/2
T2=(-1)^3*1/2^2=-1/2^2
T1+T2+T3+...................+T10
[{(1/2)-(1/2^2)}+{(1/2^3)-(1/2^4)}+....................+{(1/2^9)-(1/2^10)}]
= (1/2^2)+(1/2^4)+(1/2^6)+(1/2^8)+(1/2^10)
USING GP FORMULA
= (1/2^2)(1-(1/2^2)^5)/(1-(1/2^2))
=(1/4)*(2^10-1)*4/(2^10*3)
=(2^10-1)/(2^10*3)
=(1/3)*(1-(1/2^10))
=~ 1/3 = .333 (1/1024 Is negligible w.r.t two decimal place)
=.33
= between .25 and .5

Hi Sanjoy,

if T1 = 1/2, T2 = -1/4, T3 = 1/8, T4 = -1/16....

Is it not a gp with a multiplier of -1/2.

Using formula for GP

T = 1/2(1- (-1/2)^10)/(1-(-1/2))
= 1/2(1- (-1/2)^10)/(3/2))
= 1/3(1 - (-1/2)10)
= ~1/3

Isnt this approach also correct

Hi Mission 2012

yeah ..your approach is also correct..

Basically..In finite GP series..there is two formula

1) Sum= a(1-r^n)/(1-r) where |r|<1
2) sum= a(r^n-1)/(r-1) where |r|>1
for your case it first one..where r=-1/2