Question

se dau termeni 3;7;11;15...afla suma primilor 40 de termeni

Answer 1

$3+7+11+15+... \overset{(*)}{=}\quad (nu stim care e ultimul termen, dar il aflam) \\ \\\overset{(*)}{=}\underset{2:2=1}{(1+2)}+\underset{4:2=2}{(3+4)}+\underset{6:2=3}{(5+6)}+\underset{8:2 = 4}{(7+8)} +....+\underset{80:2 = 40}{(79+80)} = \\ \\ = 1+2+3+...+80 =\\ \\ = \dfrac{80\cdot (80+1)}{2} = \\ \\ =\dfrac{80\cdot 81}{2}= \\ \\ = 40\cdot 81 = \\ \\ =3240$

Answer 2

$a_1=3, a_2=7 =\ \textgreater \ r=a_2-a_1=7-3=\ \textgreater \ r=4$

$S_{40} = \frac{(a_1+a_{40}).40 }{2} = \frac{(3+159).40}{2} = \frac{162X40}{2} =3240$


$a_{40}=159$