Group Sequence

13 groups starting with 1 term in first 2 in second and so on gives 91 terms....so 92nd is 1/14 and 100th is 9/14. So a =9,b=14 Now before the group containing terms of the form x/17 there are 16 groups containing 16*17/2 terms=136.So 137th term is 1/17 and so 144th term is 8/17,hence c=144 So a+b+c=167

The sequence

$1 ,\quad \frac { 1 }{ 2 } ,\quad 1 ,\quad \frac { 1 }{ 3 } ,\quad \frac { 2 }{ 3 } ,\quad 1 ,\quad \frac { 1 }{ 4 } ,\quad \frac { 1 }{ 2 } ,\quad \frac { 3 }{ 4 } ,\quad 1 ,\quad \frac { 1 }{ 5 } ,\quad \frac { 2 }{ 5 } ,\quad \cdots \cdots$

Can be written as this

$\frac { 1 }{ 1 } ,\quad \frac { 1 }{ 2 } ,\quad \frac { 2 }{ 2 } ,\quad \frac { 1 }{ 3 } ,\quad \frac { 2 }{ 3 } ,\quad \frac { 3 }{ 3 } ,\quad \frac { 1 }{ 4 } ,\quad \frac { 2 }{ 4 } ,\quad \frac { 3 }{ 4 } ,\quad \frac { 4 }{ 4 } ,\quad \frac { 1 }{ 5 } ,\quad \frac { 2 }{ 5 } ,\quad \cdots \cdots$

If we were to group them, this occurs:

$\left\{ \frac { 1 }{ 1 } \right\} ,\quad \left\{ \frac { 1 }{ 2 } ,\quad \frac { 2 }{ 2 } \right\} ,\quad \left\{ \frac { 1 }{ 3 } ,\quad \frac { 2 }{ 3 } ,\quad \frac { 3 }{ 3 } \right\} ,\quad \left\{ \frac { 1 }{ 4 } ,\quad \frac { 2 }{ 4 } ,\quad \frac { 3 }{ 4 } ,\quad \frac { 4 }{ 4 } \right\} ,\quad \left\{ \frac { 1 }{ 5 } ,\quad \frac { 2 }{ 5 } ,\quad \cdots \cdots \right\} \quad \cdots \cdots$

We can clearly see it's pattern: that the kth term in nth group can be expressed as $\frac { k }{ n }$ .

Therefore, the second question regarding $\frac { 8 }{ 17 }$ , we can see that $\frac { 8 }{ 17 }$ is located at 8th term in 17th group.

Notice that the very last term's location of nth group in whole sequence can be expressed as $\frac { n(n+1) }{ 2 }$ . This is due to the fact that it is $1 + 2 + 3 + 4 + ... + n = \sum _{ k=1 }^{ n }{ k }$ . For example, the last term's location of group 3 in whole sequence is 6th term. In other words, $\frac { 3(3+1) }{ 2 } =6$ .

Therefore, last term's location of 16th group can be calculated, $\frac { 16(16+1) }{ 2 } =136$ . This means that starting point of group 17 is 137th term, which would be $\frac { 1 }{ 17 }$ .

$\frac { 8 }{ 17 }$ would be located 7 terms away. Therefore, the location of $\frac { 8 }{ 17 }$ in the sequence will be 137 + 7 = $\boxed{144}$

For 100th term, we must calculate which group does 100th term belong to.

We can do this by finding positive integer n that approximates $\frac { n(n+1) }{ 2 } = 100$ .

Doing a bit of algebra, we calculate n to be about 13.65. Since n has to be integer part of this, n is 13. This means that 100th term must be located in 14th group.

$\frac { 13(13+1) }{ 2 } =91$ . Therefore 13th group finishes at 91th term. Which means that 14th group starts from 92th term, which would be $\frac { 1 }{ 14 }$ . And therefore 100th term would be $\frac { 9 }{ 14 }$ .

Therefore we calculate a, b, and c to be $\boxed{a=9}$ , $\boxed{b=14}$ , $\boxed{c=144}$ respectively. Sum them all, we get $\boxed{167}$ .