Question

Sa se calculeze suma primilor 20 de termeni ai progresiei aritmetice (an)n>=1, stiind ca a4-a2=4 si a1+a3+a5+a6=30

Answer 1

$\displaystyle a_4-a_2=4 \\ a_{4-1}+r-(a_{2-1}+r)=4 \\ a_3+r-(a_1+r)=4 \\ a_1+3r-a_1-r=4 \\ 3r-r=4 \\ 2r=4 \Rightarrow r= \frac{4}{2} \Rightarrow r=2 \\ a_1+a_3+a_5+a_6=30 \\ a_1+a_{3-1}+r+a_{5-1}+r+a_{6-1}+r=30 \\ a_1+a_2+r+a_4+r+a_5+r=30 \\ a_1+a_1+2r+a_1+4r+a_1+5r=30 \\ 4a_1+11r=30 \\ 4a_1+11 \cdot 2=30 \\ 4a_1+22=30 \\ 4a_1=30-22 \\ 4a_1=8 \Rightarrow a_1= \frac{8}{4} \Rightarrow a_1=2$
$\displaystyle S_{20}= \frac{2 \cdot 2+(20-1) \cdot 2}{2} \cdot 20 \\ S_{20}=(4+19 \cdot 2) \cdot 10 \\ S_{20}=(4+38) \cdot 10 \\ S_{20}=42 \cdot 10 \\ S_{20}=420$