praffulpatel wrote:

For integers n from 1 to 11, inclusive, the nth term of a certain sequence is given by

Attachment:

Q.png

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If S is the sum of the first 11 terms in the sequence then S is

(A) Greater than 2

(B) Between 1 and 2

(C) Between 0 and 1/2

(D) Between -1/2 and 0

(E) Less than -1/2

I got the answer correct but i am looking for the easiest and fastest method to solve this.

Method I :Crude

The fastest way is to realize that

1. The sign of the term with the greatest absolute value( that is the first term) has a negative sign attached to it

2. The given range is all within the precision of 1 decimal value,i.e. between 0.0 and -0.5 or less than -0.5

3.The greater the denominator, the more will be the no of decimal places

Take the sum of the first two terms : \(-\frac{1}{3}+\frac{1}{9} = \frac{1-3}{9} = \frac{-2}{9} = -2*0.11(repeating) = -0.222222....\)

Rest of the remaining terms, can ONLY change the above value after the first decimal place. Thus, the value would hover around -0.2XXXX and thus the answer is D.

Method II :

A little observation and fair pairing of the terms would show that the sum of the first two terms : \(\frac{-2}{3^2}\) , the pair of the 3rd and 4th terms : \(\frac{-2}{3^4}\) and so on.

Take an infinite sum of the series with terms as \(\frac{-2}{3^2} , \frac{-2}{3^4},.. \to -2*(\frac{1}{3^2}+\frac{1}{3^4}+...) = -2*\frac{\frac{1}{3^2}}{1-\frac{1}{9}\) = \(-2*\frac{1}{8} = -0.25\). Thus, the given series can never sum up-to a value which is more than -0.25.

D.

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