Question

a)a3=3;a7=11;a15=? ÷
b)a5=4;a8=10;a19=? ÷
c)a2=3;a9=13;a17=? ÷

Answer 1

a)a3=3; a7=11 a15=?
a3=a1+2r. ;
a7=a1+6r
se formeaza un sistem: a1+2r=3 |•(-1)
a1+6r=11
=> -2r+6r=-3+11 => 4r=8 |:4 ----->r=2
Am aflat ca ratia este 2.
Aflam a1: a1+4=3--->a1= -1
a15=a1+14r----> a15= -1+28=27--> a15=27
b) a5=a1+4r
a8=a1+7r. iar se formeaza sistem: a1+4r=4 | •(-1)
a1+7r=10
3r=6 |:3---> r=2
Aflam a1: a1+4*2=4=> a1+8=4--+>a1= -4
a19=a1+18r => -4+ 18*2---> a19= 32
c)a1+r=3 |•(-1)
a1+8r=13
=> 7r=10---> r=10/7
Aflam a1: a1+10/7=3---->a1=11/7
a17=a1+16r => a17= 11/7+ 16*10/7---> a17= 171/7--> a17=171/7

Answer 2

$\displaystyle a).a_3=3,~a_7=11,~a_{15}=?\\a_3=3 \Rightarrow a_{3-1}+r=3 \Rightarrow a_2+r=3\Rightarrow a_1+2r=3\Rightarrow a_1=3-2r\\a_7=11\Rightarrow a_{7-1}+r=11 \Rightarrow a_6+r=11 \Rightarrow a_1+6r=11 \Rightarrow \\ \Rightarrow 3-2r+6r=11 \Rightarrow -2r+6r=11-3 \Rightarrow 4r=8 \Rightarrow r= \frac{8}{4} \Rightarrow r=2 \\ a_1=3-2r \Rightarrow a_1=3-2\cdot2 \Rightarrow a_1=3-4\Rightarrow a_1=-1$
$\displaystyle a_{15}=a_{15-1}+r \Rightarrow a_{15}=a_{14}+r \Rightarrow a_{15}=a_1+14r \Rightarrow \\ \Rightarrow a_{15}=-1+14 \cdot 2 \Rightarrow a_{15}=-1+28 \Rightarrow \boxed{a_{15}=27}$
$\displaystyle b).a_5=4 ,~a_8=10,~a_{19}=? \\ a_5=4 \Rightarrow a_{5-1}+r=4 \Rightarrow a_4+r=4 \Rightarrow a_1+4r=4 \Rightarrow a_1=4-4r \\ a_8=10 \Rightarrow a_{8-1}+r=10 \Rightarrow a_7+r=10 \Rightarrow a_1+7r=10 \Rightarrow \\ \Rightarrow 4-4r+7r=10 \Rightarrow -4r+7r=10-4 \Rightarrow 3r=6 \Rightarrow r= \frac{6}{3} \Rightarrow r=2 \\ a_1=4-4r \Rightarrow a_1=4-4 \cdot 2 \Rightarrow a_1=4-8 \Rightarrow a_1=-4$
$\displaystyle a_{19}=a_{19-1}+r \Rightarrow a_{19}=a_{18}+r \Rightarrow a_{19}=a_1+18r \Rightarrow \\ \Rightarrow a_{19}=-4+18 \cdot 2 \Rightarrow a_{19}=-4+36 \Rightarrow \boxed{a_{19}=32}$
$\displaystyle c).a_2=3 ,~a_9=13,~a_{17}=? \\ a_2=3 \Rightarrow a_{2-1}+r=3 \Rightarrow a_1+r=3 \Rightarrow a_1=3-r \\ a_9=13 \Rightarrow a_{9-1}+r=13 \Rightarrow a_8+r=13 \Rightarrow a_1+8r=13 \Rightarrow \\ \Rightarrow 3-r+8r=13 \Rightarrow -r+8r=13-3 \Rightarrow 7r=10 \Rightarrow r= \frac{10}{7} \\ a_1=3-r \Rightarrow a_1=3- \frac{10}{7} \Rightarrow a_1= \frac{21-10}{7} \Rightarrow a_1= \frac{11}{7}$
$\displaystyle a_{17}=a_{17-1}+r \Rightarrow a_{17}=a_{16}+r \Rightarrow a_{17}=a_1+16r \Rightarrow \\ \Rightarrow a_{17}= \frac{11}{7} +16 \cdot \frac{10}{7} \Rightarrow a_{17}= \frac{11}{7} + \frac{160}{7} \Rightarrow \boxed{a_{17}= \frac{171}{7} }$