Explanation

Remember, the key to most GRE sequence questions is discovering a pattern among the terms. So start off by finding the first few terms and looking for a pattern.

\(a1 = 1 -\frac{1}{1+1}= 1-\frac{1}{2}\).
\(a2 = \frac{1}{2} -\frac{1}{2+1}= \frac{1}{2} - \frac{1}{3}\).
\(a3 = \frac{1}{3} -\frac{1}{3+1}= \frac{1}{3} - \frac{1}{4}\).

Hence sum \(a1 + a2 + a3 ....... = 1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ......\)

So alternate terms terms cancel out. So

\(a100 = \frac{1}{100} - \frac{1}{101}\)

So \(a1 + a2 + a3 ...... a100\)=\(1 - \frac{1}{101} = \frac{100}{101}\).


Hence option D is correct.
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Why don't we get the correct answer for this using the sum of n terms of AP formula?

I tried: \(K1 = 0.5; K100= 0.000099; Average of K1 and K100 = 0.2500495; No. of terms in AP = 100; So, sum should be 25.\)

They are not in A.P dear!

\(a_1 = \frac{1}{2}\)

\(a_2 = \frac{1}{6}\)

\(a_3 = \frac{1}{12}\)

Common difference (d) = \(a_2 - a_1 = a_3 - a_2\)

\(a_2 - a_1 = \frac{-1}{3}\)
\(a_3 - a_2 = \frac{-1}{12}\)
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Thank you! What kind of progression would this be called?

It is called as Telescoping Series
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Wow. Every day learning something new. Thanks for your help!

it is relevant to know for the GRE ?

I mean the question could be solved without knowing this
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Totally, we don't need to know it
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