Question

VA ROG SA MA AJUTATI LA CELE 4 EXERCITII!!!!!Macar 2 exercitii din 4.....

Answer 1

Ai  o  PA  de  ratir  r=7-1=6
termenul  general
an=a1+(n-1)*r  n=10
a10=1+9*6=55
3)Ai  o  progresie  aritmetica  cu 9  termeni
Ratia  r=5-1=4
aplici  formula  sumei
Sn=(a1+an)*n/2=(1+25)*9/2=117
4)Formula  termenului  general  unei  PG
an=a1*q^(n-1)
a1=243*(1/3)^(n-1)=3^5*(1/3)^8=
3^5*3^(-8)=3^(5-8)=3^(-³)=1/27

Answer 2

1)
$a_n=6n-5\\a_1_0=6*10-5=60-5=55$
3)
Pentru ca 3 numere a,b,c sa fie intr-o progresi artimetica
[tex]b=\frac{a+c}{2}\\2*3^{x+1}=3^{x-1}+5*{3^{x}}+1\\ 6*3^{x}=\frac{3^{x}}{3}+5*3^{x}+1\\Notam\ 3^{x}=t\\ 6t=\frac{t}{3}+5t+1\\\frac{t}{3}+\frac{15t}{3}+\frac{3}{3}=6t\\ \frac{t+15t+3}{3}=6t\\16t+3=6t*3\\16t+3=18t\\18t-16t=3\\2t=3\\t=\frac{3}{2} \\3^{x}=\frac{3}{2}\\
4)
ratia=4\\
S=1+5+9+....+25\\
S=25+21+...+5+1\\
2S=26+26+...+26\\
Numarul termenilor=(25-1):4+1=7\\
2S=26*7\\
S=26*7:2\\
S=13*7\\
S=91\\
5)
[tex]b_n=b_n_-_1*q\\b_2=243*\frac{1}{3}=81\\b_3=\frac{81}{3}=27\\ b_9=\frac{1}{9}[/tex]