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Aye can u help me? :)
Given the following sequence, find the 22nd term:

10.5, 11, 11.5, 12, 12.5, . . .

A. 20.5
B. 21
C. 21.5
D. 22

Respuesta :

LammettHash
The given sequence is arithmetic with a common difference of [tex]\dfrac12[/tex] between each term.

Recursively, you can define the sequence as

[tex]a_n=a_{n-1}+\dfrac12[/tex]

and solve explicitly for the [tex]n[/tex]th term by recursively substituting the right hand side:

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

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

[tex]a_n=\dfrac{21}2+\dfrac{n-1}2[/tex]

So, the 22nd term in the sequence is

[tex]a_{22}=\dfrac{21}2+\dfrac{22-1}2=21[/tex]

Answer:

the answer is B.

Step-by-step explanation:

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