Question

First term of $11th$term of the following groups $\left(1\right),(2,3,4),(5,6,7,8,9),...$is

• A. $89$
• B. $97$
• C. $101$
• D. $123$

Answer 1

The correct option is C

$101$

Step 1: Find the pattern in given groups.

We have been given the groups $\left(1\right),(2,3,4),(5,6,7,8,9),...$

We need to find the first of the $11th$term of given sequence of group.

Consider the group $(2,3,4)$

We can observe that the elements of this group are in A.P. with a common difference of $1$.

Similarly consider a group $(5,6,7,8,9)$

We can observe that the elements of this group are in A.P. with a common difference of $1$

This means, in each group, the numbers are in A.P. with a common difference of $1$

Step 2: Find the sequence of the first terms of each group.

The sequence of the first terms of each group is $1,2,5,10,...,t_n$

Let $P$ be the sum of all terms of the above sequence.

$$\begin{gathered} \Rightarrow P=1+2+5+10+...+t_n+0.............\left(i\right) \\ \Rightarrow P=0+1+2+5+10+...+t_n.............\left(ii\right) \end{gathered}$$

Subtracting $\left(i\right)-\left(ii\right)$,

$$\begin{gathered} \Rightarrow 0=1+[1+3+5+10+....(n-1\left)terms\right]-t_n \\ \Rightarrow t_n=1+[1+3+5+10+....(n-1\left)terms\right].............\left(iii\right) \end{gathered}$$

Step 3: Find the sum $1+3+5+10+....(n-1)terms$

If we observe the sum, $1+3+5+10+....(n-1)terms$

The terms are in A.P. with common difference $d=2$

We know, the formula for the sum of first $\text{'}n\text{'}$terms of A.P. is,

${\mathit{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{[}\mathbf{2}\mathit{a}\mathbf{+}\mathbf{(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{\left)}\mathit{d}\mathbf{\right]}$

So, the sum $1+3+5+10+....(n-1)terms$ would be,

$$\begin{gathered} \Rightarrow S_{n-1}=\frac{n-1}{2}\left[2\right(1)+((n-1)-1)\times 2] \\ \Rightarrow S_{n-1}=\frac{n-1}{2}[2+(n-1-1)\times 2] \\ \Rightarrow S_{n-1}=\frac{n-1}{2}[2+(n-2)\times 2] \\ \Rightarrow S_{n-1}=\frac{n-1}{2}[2+2n-4] \\ \Rightarrow S_{n-1}=\frac{n-1}{2}[2n-2] \\ \Rightarrow S_{n-1}=(n-1)\times (n-1) \\ \Rightarrow S_{n-1}={(n-1)}^2 \end{gathered}$$

Step 4: Find the formula of $t_n$by solving an equation $\left(iii\right)$

$$\begin{gathered} \Rightarrow t_n=1+[1+3+5+10+....(n-1\left)terms\right] \\ \Rightarrow t_n=1+S_{n-1} \\ \Rightarrow t_n=1+{(n-1)}^2 \end{gathered}$$

Step 5: Find the required term using the above formula of $t_n$.

the first the $11th$term of a given sequence of groups would be,

$$\begin{gathered} \Rightarrow t_n=1+{(n-1)}^2 \\ \Rightarrow t_{11}=1+{(11-1)}^2 \\ \Rightarrow t_{11}=1+{10}^2 \\ \Rightarrow t_{11}=1+100 \\ \Rightarrow t_{11}=101 \end{gathered}$$

Therefore, option (C) is the correct answer.