Question

Consider the sequence of numbers $2,5,5,8,8,8,11,11,11,11,....$ The ${150}^{th}$ term of the sequence is

• A. $48$
• B. $50$
• C. $47$
• D. $53$

Answer 1

The correct option is B $50$

$2$ is coming one time

$5$ is coming two times

$8$ is coming four times and so on.

We can see that $2$ is $1^{st}$ term, $5$ is $2^{nd}$ term, $8$ is $4^{th}$ term, $11$ is $7^{th}$ term and so on.

Hence, $(3k+2)$ is the ${(\frac{k(k+1)}{2}+1)}^{th}$ term, where $k\in I$

Take $k=16\Rightarrow (\frac{k(k+1)}{2}+1)=137k=17\Rightarrow (\frac{k(k+1)}{2}+1)=154$

Hence, ${150}^{th}$ term will be for $k=16$

The number will be $3k+2=3\times 16+2=50$

Hence, ${150}^{th}$ term is $50$