Question

The first term of ${11}^{th}$ group in the following groups $\left(1\right),(2,3,4),(5,6,7,8,9),.....$ is

• A. $97$
• B. $101$
• C. $98$
• D. $123$

Answer 1

The correct option is B $101$

The number in the $n^{th}$ group are in $A.P.$

With common difference$1.$

$\therefore$ The First in each group are the sequence$1,2,5,10,\dots \dots tn$

Let$S=1+2+5+10+\dots \dots +t_n+0$ ...... $\left(1\right)$

$S=0+1+2+5+10\dots \dots +t_n$ …… $\left(2\right)$

Subtracting $\left(1\right)–\left(2\right)$ and we get.

$0=1+(1+3+5+\dots \dots (n-1)terms)–t_n$

$\Rightarrow t_n=1+(1+3+5+\dots \dots (n-1))$terms

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

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

$={(n-1)}^2$

$\Rightarrow t_n=1+\frac{n-1}{2}(2+(n-2\left)2\right)$

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

$\Rightarrow tn=n^2-2n+2$

$\text{at}n=11$

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

$t_{11}=101$