Question

The sequence $a_1,a_2,a_3,....,a_n,...$ is such that $a_n=\frac{a_{n-1}+a_{n-2}}{2}$ for all $n\ge 3$. If $a_3=4$ and $a_5=20$, what is the value of $a_6$?

• A. $12$
• B. $16$
• C. $20$
• D. $24$
• E. $28$

Answer 1

The correct option is E $28$

According to this formula, it is necessary to know the two prior terms in the sequence to determine the value of a term; that is, it is necessary to know both $a_{n-1}$ and $a_{n-2}$ to find $a_n$.

Therefore, to find $a_6$, the values of as and $a_4$ must be determined. To find $a_4$, let $a_n=a_5$, which makes $a_{n-1}=a$, and $a_{n-2}=a_3$. Then, by substituting the given values into the formula

$a_n=\frac{a_{n-1}+a_{n-2}}{2}$

$a_5=\frac{a_4+a_3}{2}$

$20=\frac{a_4+4}{2}$ substitute known values

$40=a_4+4$ multiply both sides

$36=a_4$ subtract $4$ from both sides

Then, letting $a_n=a_6$, substitute the known values:

$a_6=\frac{a_5+a_4}{2}$

$a_6=\frac{20+36}{2}$ substitute known values

$a_6=\frac{56}{2}$ simplify

$a_6=28$

The correct answer is E.