Question

find the seventh term of the sequence whose first two terms are 2 and 3 and each term beginning with the third is the sum of the two terms preceding it.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{two}\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{series}\mathrm{be}\mathrm{given}\mathrm{by}: \\ {\mathrm{a}}_0=2 \\ {\mathrm{a}}_1=3 \\ \mathrm{Since}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that},\mathrm{each}\mathrm{term}\mathrm{beginning}\mathrm{with}\mathrm{the}\mathrm{third}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{two}\mathrm{preceding}\mathrm{terms},\mathrm{thus}: \\ {\mathrm{a}}_2={\mathrm{a}}_0+{\mathrm{a}}_1=2+3=5 \\ {\mathrm{a}}_3={\mathrm{a}}_1+{\mathrm{a}}_2=3+5=8 \\ {\mathrm{a}}_4={\mathrm{a}}_2+{\mathrm{a}}_3=5+8=13 \\ {\mathrm{a}}_5={\mathrm{a}}_3+{\mathrm{a}}_4=8+13=21 \\ {\mathrm{a}}_6={\mathrm{a}}_4+{\mathrm{a}}_5=13+21=34 \\ \mathrm{So},\mathrm{the}\mathrm{seventh}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{series}\mathrm{will}\mathrm{be}34. \end{gathered}$$