Question

Aye can u help me? :)

Answer 1

The given sequence is arithmetic with a common difference of $\dfrac12$ between each term.

Recursively, you can define the sequence as

$a_n=a_{n-1}+\dfrac12$

and solve explicitly for the $n$th term by recursively substituting the right hand side:

$a_n=a_{n-1}+\dfrac12$
$\implies~a_n=a_{n-2}+2\times\dfrac12$
$\implies~a_n=a_{n-3}+3\times\dfrac12$
$\implies\cdots\implies~a_n=a_1+(n-1)\dfrac12$

The first term of the sequence is $a_1=\dfrac{21}2$, so the $n$th term is given by the formula

$a_n=\dfrac{21}2+\dfrac{n-1}2$

So, the 22nd term in the sequence is

$a_{22}=\dfrac{21}2+\dfrac{22-1}2=21$

Answer 2

Answer:

the answer is B.

Step-by-step explanation: