Question

Va rog mult sa ma ajutati si pe mine la ex 1!!

Answer 1

Răspuns:

420

Explicație pas cu pas:

$a_{4} - a_{2} = 4 \\ a_{1} + 3\cdot r - a_{1} - r = 4 \\ 2\cdot r = 4 \implies \bf r = 2$

$a_{1} + a_{3} + a_{5} + a_{6} = 30$

$a_{1} + a_{1} + 2 \cdot r + a_{1} + 4 \cdot r + a_{1} + 5 \cdot r = 30$

$4 \cdot a_{1} + 11 \cdot 2 = 30 \iff 4 \cdot a_{1} = 30 - 22 \\ a_{1} = 8 : 2 \implies \bf a_{1} = 2$

$a_{20} = a_{1} + 19 \cdot r = 2 + 19 \cdot 2 = 2 + 38 = 40$

$S_{n} = \frac{(a_{1} + a_{n})\cdot n}{2}$

$S_{20} = \frac{(2 + 40)\cdot 20}{2} = \bf 420$

Answer 2

$\it Fie\ \ r\ -\ ra\c{\it t}ia\ \ progresiei\\ \\ a_4-a_2=4 \Rightarrow a_2+2r-a_2=4 \Rightarrow 2r=4\Big|_{:2} \Rightarrow r=2$

$\it a_1+a_3+a_5+a_6=30 \Rightarrow a_3-2r+a_3+a_3+2r+a_3+3r=30 \Rightarrow \\ \\ 4a_3+3r=30 \Rightarrow 4a_3+3\cdot2=30 \Rightarrow 4a_3+6=30\Big|_{-6} \Rightarrow \\ \\ 4a_3=24\Big|_{:4} \Rightarrow a_3=6\\ \\ a_1=a_3-2r=6-2\cdot2=6-4=2\\ \\ a_{20}=a_1+19r=2+19\cdot2=2+38=40$

$\it S_{20}=\dfrac{(a_1+a_{20})\cdot20}{2}=(2+40)\cdot10=420$