2018-II-12

Ans: A
Consider

$\begin{array}{rcl}{a}_3& =& a_1+a_2\\ 21& =& a_1+a_2\\ a_1& =& 21–a_2\dots \text{①}\end{array}$

Consider

$\begin{array}{rcl}{a}_4& =& a_2+a_3\\ a_4& =& a_2+21\\ a_2& =& a_4–21\dots \text{②}\end{array}$

Sub. $\text{②}$ into $\text{①}$, we have

$\begin{array}{rcl}{a}_1& =& 21–(a_4–21)\\ a_1& =& 42–a_4\dots \text{③}\end{array}$

Consider

$\begin{array}{rcl}{a}_5& =& a_3+a_4\\ a_5& =& 21+a_4\\ a_5–a_4& =& 21\dots \text{④}\end{array}$

Consider

$\begin{array}{rcl}{a}_6& =& a_4+a_5\\ a_4+a_5& =& 89\dots \text{⑤}\end{array}$

$\text{⑤}–\text{④}$, we have

$\begin{array}{rcl}2a_4& =& 68\\ a_4& =& 34\end{array}$

Sub. $a_4=34$ into $\text{③}$, we have

$\begin{array}{rcl}{a}_1& =& 42–34\\ a_1& =& 8\end{array}$