Question

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

• A. 179
• B. 99
• C. 79
• D. 59

Answer 1

Answer: (A)

$n\ge 3,a_3=\frac{a_1+a_2}{2}$...(1)
$a_4=\frac{\left(a_1+a_2\right)+a_3}{3}=\frac{2a_3+a_3}{3}\Rightarrow a_4=a_3$
$a_5=\frac{\left(a_1+a_2+a_3+a_4\right)}{4}=\frac{3a_4+a_4}{4}=a_4$
$a_3=a_4=a_5=\dots \dots \dots =a_9=99$
put in equation ( 1 )
$99=\frac{19+a_2}{2}\Rightarrow a_2=179$