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Given the following sequence, find the 22nd term:


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

A20.5
B21
C21.5
D22

Verified Answer :

LammettHash

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:

the answer is B.

Step-by-step explanation:

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