Question

First term of $11th$ term of the following groups $(1),(2,3,4),(5,6,7,8,9).....$ is
(1) $89$
(2) $97$
(3) $101$
(4) $123$

Answer 1

Hint:The groups of the given series are in Arithmetic progression. So, to find the First term of $11th$ term of the group $(1),(2,3,4),(5,6,7,8,9).....$ use the formula of A.P. The common difference between the first term of each group is $1$.

Formula Used:
Sum of $n$ terms of A.P. (Arithmetic Progression) Formula –
${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
Here, $a$ is the first term of the sequence. Also, the common difference between the terms is $d$.

Complete step by step Solution:
Given that,
The sequence is $(1),(2,3,4),(5,6,7,8,9).....$
The sequence of first term of each group is \[1,2,5,10,......,{p_n}\]
Let, $S$be the sum of the first term of each group
$ \Rightarrow S = 1 + 2 + 5 + 10 + ...... + {p_n} - - - - - - - - - (1)$
Also, $S = 0 + 1 + 2 + 5 + 10 + ...... + {p_n} - - - - - - - - - (2)$
Subtracting equation (2) from (1)
$0 = 1 + \left( {1 + 3 + 5 + \ldots \left( {n - 1} \right){\text{ }}terms} \right){\text{ }}-{p_n}$
$ \Rightarrow {p_n} = 1 + \left( {1 + 3 + 5 + \ldots \left( {n - 1} \right){\text{ }}terms} \right) - - - - - - (3)$
The sequence $1 + 3 + 5 + \ldots \left( {n - 1} \right){\text{ }}terms$is in A.P.
Here, $a = 1,d = 2,n = n - 1$
Using the sum of $n$ terms of A.P. Formula
${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
${S_n} = \dfrac{{(n - 1)}}{2}\left[ {2(1) + ((n - 1) - 1)2} \right]$
${S_n} = \dfrac{{(n - 1)}}{2}\left[ {2 + (n - 2)2} \right]$
${S_n} = \dfrac{{(n - 1)}}{2}\left[ {2n - 2} \right]$
${S_n} = {(n - 1)^2}$
Equation (3) will be
${p_{11}} = 1 + {(11 - 1)^2}$
Put $n = 11$ in above equation,
${p_{11}} = 1 + {(11 - 1)^2}$
${p_{11}} = 1 + 100$
${p_{11}} = 101$

Hence, the correct option is 3.

Note:To know a common difference check the sequence carefully. Observe the number of elements of the consecutive groups which form an A.P. carefully. When the equation will be formed it will get easier to find the value of the first term of $11th$ group.