Question

1. calculati suma 1+5+9+13...25
2. caculati suma 1+3+5+...21
2.fie progresia aritmetica in care a3=5,a6=11.calculati a9
3.calculati suma 1+11+21+31 +...+110
4.calculati al cincilea termen al unei progresii aritmetice,stiind ca primul termen al progresiei este 7 si al doile nr. este 9
5.sa se determine al zecelea termen al sirului 1,7,13,19
6. sa se determine al cincisprezecelea termen al sirului -5,-2,1 4,...
7.sa se calculeze suma primilor 10 termeni ai unei progresii aritmetice in care a1=3 si r=2
8.sa se calculeze produsul primilor cinci termeni a1=-1 si r=4
9. sa se determine nr. realx,stiind ca x-2,x si 2x+4 sunt in progresie aritmetica

Answer 1

$\displaystyle 1).1+5+9+13+...+25 \\ 25=1+(n-1) \cdot 4\Rightarrow 25=1+4n-4 \Rightarrow 4n=25-1+4 \Rightarrow \\ \Rightarrow 4n=28 \Rightarrow n= \frac{28}{4} \Rightarrow n=7 \\ S_7= \frac{2+6 \cdot 4}{2} \cdot 7 \\ \\ S_7= \frac{2+24}{2} \cdot 7 \\ \\ S_7= \frac{26}{2} \cdot 7 \\ \\ S_7=13 \cdot 7 \\ S_7=91$
$\displaystyle 2).1+3+5+...+21 \\ 21=1+(n-1) \cdot 2 \Rightarrow 21=1+2n-2 \Rightarrow 2n=21-1+2 \Rightarrow \\ \Rightarrow 2n=22 \Rightarrow n= \frac{22}{2} \Rightarrow n=11 \\ S_{11}= \frac{2+10 \cdot 2}{2} \cdot 11 \\ \\ S_{11}= \frac{2+20}{2} \cdot 11 \\ \\ S_{11}= \frac{22}{2} \cdot 11 \\ \\ S_{11}=11 \cdot 11 \\ S_{11}=121$
$\displaystyle 3).a_3=5,~a_6=11,~a_9=? \\ a_3=5 \Rightarrow a_{3-1}+r=5 \Rightarrow a_2+r=5 \Rightarrow a_1+2r=5 \Rightarrow a_1=5-2r \\ a_6=11 \Rightarrow a_{6-1}+r=11 \Rightarrow a_5+r=11 \Rightarrow a_1+5r=11 \Rightarrow \\ \Rightarrow 5-2r+5r=11 \Rightarrow 5+3r=11 \Rightarrow 3r=11-5 \Rightarrow 3r=6 \Rightarrow \\ \Rightarrow r= \frac{6}{3} \Rightarrow r=2 \\ a_1=5-2r \Rightarrow a_1=5-2 \cdot 2 \Rightarrow a_1=5-4 \Rightarrow a_1=1$
$\displaystyle a_9=a_{9-1}+r \Rightarrow a_9=a_8+r \Rightarrow a_9=a_1+8r \Rightarrow a_9=1+8 \cdot 2 \Rightarrow \\ \Rightarrow a_9=1+16 \Rightarrow a_9=17$
$\displaystyle 4).1+11+21+31+...+111 \\ 111=1+(n-1) \cdot 10 \Rightarrow 111=1+10n-10 \Rightarrow 10n=111-1+10 \Rightarrow \\ \Rightarrow 10n=120 \Rightarrow n= \frac{120}{10} \Rightarrow n=12 \\ S_{12}= \frac{2+11 \cdot 10}{2} \cdot 12 \\ \\ S_{12}= \frac{2+110}{2} \cdot 12 \\ \\ S_{12}= \frac{112}{2} \cdot 12 \\ \\ S_{12}=56 \cdot 12 \\ S_{12}=672$
$\displaystyle 5).a_1=7,~a_2=9 \\ a_2=9 \Rightarrow a_{2-1}+r=9 \Rightarrow a_1+r=9 \Rightarrow 7+r=9 \Rightarrow \\ \Rightarrow r=9-7 \Rightarrow r=2 \\ a_5=a_{5-1}+r \Rightarrow a_5=a_4+r \Rightarrow a_5=a_1+4r \Rightarrow a_5=7+4 \cdot 2 \Rightarrow \\ \Rightarrow a_5=7+8 \Rightarrow a_5=15$
$\displaystyle 6).1,7,13,19 \\ a_1=1,~a_2=7,~a_3=13,~a_4=19 \\ r=7-1 \Rightarrow r=6 \\ a_1_0=a_{10-1}+r \Rightarrow a_{10}=a_9+r \Rightarrow a_{10}=a_1+9r \Rightarrow a_{10}=1+9 \cdot 6 \Rightarrow \\ \Rightarrow a_{10}=1+54 \Rightarrow a_{10}=55$
$\displaystyle 7).-5,-2,1,4 \\ a_1=-5,~a_2=-2,~a_3=1,~a_4=4 \\ r=-2-(-5) \Rightarrow r=-2+5 \Rightarrow r=3 \\ a_{15}=a_{15-1}+r \Rightarrow a_{15}=a_{14}+r \Rightarrow a_{15}=a_1+14r \Rightarrow \\ \Rightarrow a_{15}=-5+14 \cdot 3 \Rightarrow a_{15}=-5+42 \Rightarrow a_{15}=37$
$\displaystyle 8).a_1=3,~r=2,~S_{10}=? \\ S_{10}= \frac{6+9 \cdot 2}{2} \cdot 10 \\ \\ S_{10}= \frac{6+18}{2} \cdot 10 \\ \\ S_{10}= \frac{24}{2} \cdot 10 \\ \\ S_{10}=12 \cdot 10 \\ S_{10}=120$
$\displaystyle 9).a_1=-1,~r=4 \\ a_2=a_{2-1}+r \Rightarrow a_2=a_1+r \Rightarrow a_2=-1+4 \Rightarrow a_2=3 \\ a_3=a_{3-1}+r \Rightarrow a_3=a_2+r \Rightarrow a_3=3+4 \Rightarrow a_3=7 \\ a_4= a_{4-1}+r \Rightarrow a_4=a_3+r \Rightarrow a_4=7+4 \Rightarrow a_4=11 \\ a_5=a_{5-1}+r \Rightarrow a_5=a_4+r \Rightarrow a_5=11+4 \Rightarrow a_5=15 \\ a_1 \cdot a_2 \cdot a_3 \cdot a_4 \cdot a_5=-1 \cdot 3 \cdot 7 \cdot 11 \cdot 15=-3465$
$\displaystyle 10).x-2,~x,~2x+4 \\ x= \frac{x-2+2x+4}{2} \Rightarrow x= \frac{3x+2}{2} \Rightarrow 2x=3x+2 \Rightarrow 2x-3x=2 \Rightarrow \\ \Rightarrow -x=2 \Rightarrow x=-2$