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**Hint:**The given groups are in A.P with common difference 1 . Use the formula of A.P to find the required answer .

**Complete step-by-step answer:**The numbers in the $n^{th}$ group are in A.P , with the common difference 1 .

The first in each group are in the sequence 1,2,5,10,………. $t_{n}$

Let $S=1+2+5+10+……+ t_{n}+0--(i)$

$S=0+1+2+5+10+……+ t_{n}--(ii)$

Subtracting (i)-(ii) and we get ,

$0=1+{1+3+5+…..(n-1)terms}- t_{n}$

$\Rightarrow t_{n}=1+{1+3+5+…..(n-1)terms}$

So, 1+3+5+…..(n-1)terms terms in sum of (n-1) terms of an A.P with d = 2

$=\dfrac {(n-1)}{2}[2 \times 1 +(n-2) \times 2]$

Now we simply the equation to get,

$=(n-1)^2$

Now we use the formula for sum of AP

$\Rightarrow t_{n}=1+\dfrac {(n-1)}{2}[2 \times 1 +(n-2) \times 2]$

On simplifying we get,

$\Rightarrow t_{n} = 1+(n-1)^2$

$\Rightarrow t_{n}= n^2-2n+2$

We take as n=11 as given in the question,

When n=11 ,

$ t_{n}= n^2-2n+2$

On putting n as 11 we get,

$\Rightarrow t_{n}= 11^2-2 \times 11+2$

Simplifying the LHS we get,

$\Rightarrow t_{n}= 121+22+2$

$\Rightarrow t_{n}= 101$

**Thus, the required option for the given sum is b) 101**

**Note:**Students often go wrong in finding the common difference for the AP. Here we should observe carefully the numbers of elements in the consecutive groups form an A.P. Once we form the equation then it’s easy to find the term of the 11th group.

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