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Raj

answered 2015-09-20 12:57:57

The series is 1, (2,2), (3,3,3), (4,4,4,4), ...

**1st** term : 1

2nd,**3rd** terms : 2

4th, 5th,**6th** terms : 3

7th, 8th, 9th,**10th** terms : 4

11th, 12th, 13th, 14th,**15th** terms : 5

Each group ends in a sequence 1,3,6,10,15, ...

This can be represented using $\dfrac{n(n+1)}{2}$ where n=1,2,3,...

With this, we can easily find out the answer.

Say,n=7.

$\dfrac{7×8}{2}=28$

This means, 28th term is 7

=> 22nd to 28th terms will be 7

Therefore, we find largest integer value of n satisfying

$\dfrac{n(n+1)}{2} \lt 100\\

n(n+1) \lt 200$

Largest integer value of n satisfying the above condition is 13

*(because 13×14=182 < 200 and if n=14, 14×15 will become greater than 200)*

For n=13, $\dfrac{n(n+1)}{2}=\dfrac{13×14}{2}=91$

i.e., 91th term is 13

92nd term is 14

=> 92nd to 105 terms are 14*(taken 14 terms from 92, including 92)*

Therefore 100th term is 14

2nd,

4th, 5th,

7th, 8th, 9th,

11th, 12th, 13th, 14th,

Each group ends in a sequence 1,3,6,10,15, ...

This can be represented using $\dfrac{n(n+1)}{2}$ where n=1,2,3,...

With this, we can easily find out the answer.

Say,n=7.

$\dfrac{7×8}{2}=28$

This means, 28th term is 7

=> 22nd to 28th terms will be 7

Therefore, we find largest integer value of n satisfying

$\dfrac{n(n+1)}{2} \lt 100\\

n(n+1) \lt 200$

Largest integer value of n satisfying the above condition is 13

For n=13, $\dfrac{n(n+1)}{2}=\dfrac{13×14}{2}=91$

i.e., 91th term is 13

92nd term is 14

=> 92nd to 105 terms are 14

Therefore 100th term is 14

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