$a_n$ and $b_n$ are sequences of arithmetic and geometric means, whose first three terms match. Given some $a_n$ values, find $a_2^2+a_3^2=1$.

Question

For $n>3$,

• $a_{n}$ is arithmetic mean of $a_1$ through $a_{n-1}$
• $b_{n}$ is geometric mean of $b_1$ through $b_{n-1}$

For $i=1,2,3$,

• ${{a}_{i}}$ = ${{b}_{i}}$

If $a_{1} = 3$, $a_{2017} = 7$, $a_{2018}=8$, then $$a_{2}^{2}+a_{3}^{2} = \text{?}$$

What is the use of $b$ here? Where is it going to be used in solution?

Answer 1

See $2a_3=a_1+a_2$, $a_2=a_1$ and $b_3=\sqrt{a_1a_2}$. We get $a_2^2+a_3^2=18$.