Question

Sa se calculeze suma primilor 20 de termeni ai unei progresi aritmetice , stiind ca a4 - a2 = 4 si a1 + a3 + a5 + a6 = 30 . MULTUMESC

Answer 1

[tex]\div \\\;\\ a_4-a_2=4 \Longrightarrow \not a_2+2r- \not a_2=4 \Longrightarrow 2r=4 \Longrightarrow r=2[/tex]

[tex]a_1+a_3+a_5+a_6 =30 \Longrightarrow a_1+a_1+2r+a_1+4r+a_1+5r = 30 \\\;\\ \Longrightarrow 4a_1+11r=30 \Longrightarrow 4a_1+11\cdot2=30 \Longrightarrow a_1= 2.[/tex]

Progresia este :
 2, 4, 6, 8, .....2n, .....

Suma primilor 20 de termeni este:

2+4+6+ ... +40 =(2+40)·20/2=420

Answer 2

$\displaystyle a_4-a_2=4 \Rightarrow a_{4-1}+r-(a_{2-1}+r)=4 \Rightarrow a_3+r-(a_1+r)=4 \Rightarrow \\ \\ \Rightarrow a_1+3r-a_1-r=4 \Rightarrow 3r-r=4 \Rightarrow 2r=4 \Rightarrow r= \frac{4}{2} \Rightarrow r=2 \\ \\ a_1+a_3+a_5+a_6=30 \Rightarrow a_1+a_{3-1}+r+a_{5-1}+r+a_{6-1}+r=30 \Rightarrow \\ \\ \Rightarrow a_1+a_2+r+a_4+r+a_5+r=30 \Rightarrow \\ \\ \Rightarrow a_1+a_1+2r+a_1+4r+a_1+5r=30 \Rightarrow 4a_1+11r=30 \Rightarrow$

$\displaystyle \Rightarrow 4a_1+11 \cdot 2=30 \Rightarrow 4a_1+22=30 \Rightarrow 4a_1=30-22 \Rightarrow \\ \\ \Rightarrow 4a_1=8 \Rightarrow a_1= \frac{8}{4} \Rightarrow a_1=2 \\ \\ S_{20}= \frac{2 \cdot 2+(20-1) \cdot 2}{\not2} \cdot \not20 \\ \\ S_{20}=(4+19 \cdot 2) \cdot 10 \\ \\ S_{20}=(4+38) \cdot 10 \\ \\ S_{20}=42 \cdot 10 \\ \\ S_{20}=420$