1. Tardigrade
  2. Question
  3. Mathematics
  4. The sequence a1, a2, a3, ldots satisfies a1=19, a9=99, and for all n ≥ 3, an is the arithmetic mean of the first (n-1) terms. Then a2 is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The sequence $a_1, a_2, a_3, \ldots$ satisfies $a_1=19, a_9=99$, and for all $n \geq 3, a_n$ is the arithmetic mean of the first $(n-1)$ terms. Then $a_2$ is equal to

Sequences and Series

Solution:

$n \geq 3, a _3=\frac{ a _1+ a _2}{2}$...(1)
$a _4=\frac{\left( a _1+ a _2\right)+ a _3}{3}=\frac{2 a _3+ a _3}{3} \Rightarrow a _4= a _3$
$a_5=\frac{\left(a_1+a_2+a_3+a_4\right)}{4}=\frac{3 a_4+a_4}{4}=a_4 $
$a_3=a_4=a_5=\ldots \ldots \ldots=a_9=99$
put in equation ( 1 )
$99=\frac{19+ a _2}{2} \Rightarrow a _2=179 $