Question

If $t_n$ denotes $n^{th}$ term of the series $2+3+6+11+18...$ ,then $t_{50}$ is

• A. $2+{49}^2$
• B. $2+{48}^2$
• C. $2+{50}^2$
• D. $2+{51}^2$

Answer 1

The correct option is A

$2+{49}^2$

Find the term of the series

As we know that, a sequence is an arrangement of any objects or a set of numbers in a particular order followed by some rule. If $a_1,a_2,a_3,a_4,\dots \dots \dots$ etc. denote the terms of a sequence, then $1,2,3,4,\dots ..$ denotes the position of the term.

If $a_1,a_2,a_3,a_4,\dots \dots \dots$ is a sequence, then the corresponding series is given by,

$S_N=a_1+a_2+a_3+..+a_N$

Given series are $2+3+6+11+18...$.

Let,

$S_{50}=2+3+6+11+18+...+t_{49}+t_{50}$ ----------- $\left(1\right)$

$S_{50}=0+2+3+6+11+18+...+t_{49}+t_{50}$ ------- $\left(2\right)$

On subtracting equation $\left(2\right)$ from $\left(1\right)$, we get,

$\Rightarrow$ $0=\left(2+1+3+5+7+...\mathrm{up}\mathrm{to}50\mathrm{terms}\right)$

$\Rightarrow$ $t_{50}=2+\left[1+3+5+7+...\mathrm{upto}49\mathrm{terms}\right]$

$\Rightarrow$ $t_{50}=2+\frac{49}{2}\left[2\times 1+\left(49-1\right)\times 2\right]$

$\Rightarrow$ $t_{50}=2+49\left(1+48\right)$

$\Rightarrow$ $t_{50}=2+{49}^2$

Hence, option $A$ is correct answer.