Question

1. Se da progresia aritmetica (an)n>=1; a1 = 4; r = 6. Termenul a2 este:

2. Se da progresia aritmetica (an)n>=1; a1 = 4; r = 6. Termenul a5 este:

3. Se da progresia aritmetica (an)n>=1; a1 = 4; r = 6. Suma S5 este:

4. Se da progresia geometrica (bn)n>=1; b1 = 5; r = 2. Termenul b2 este:

5. Se da progresia geometrica (bn)n>=1; b1 = 5; r = 2. Termenul b5 este:

6. Se da progresia geometrica (bn)n>=1; b1 = 5; r = 2. Suma S5 este:

7. Se da sirul (an)n>=1; an = (n-1)/(n-2). Termenul a8 este:

8. Se da sirul (an)n>=1; an = (-1)^n+6. Termenul a12 este:

Answer 1

$\displaystyle\bf\\\boxed{\bf progresii~~aritmetice}~.\\\\1.~\boxed{\bf a_n=a_{n-1}+r,~\forall~n\in\mathbb{N^*},~n\geq 2}~.\\\\2.~\boxed{\bf a_n=a_1+(n-1)r,~\forall~n\in\mathbb{N^*},~n\geq 2}~.\\\\3.~\boxed{\bf S_n=\frac{\Big(a_1+a_n\Big)n}{2},~\forall~n\in\mathbb{N^*},~n\geq 2}~.\\\\4.~\boxed{\bf a_n=\frac{a_{n-1}+a_{n+1}}{2}, ~\forall~n\in\mathbb{N^*},~n\geq 2}~.\\\\------------------------$

$\displaystyle\bf\\\boxed{\bf progresii~geometrice}~.\\\\1.~\boxed{\bf b_n=b_{n-1}q,~\forall~n\in\mathbb{N^*},~n\geq 2,~q\neq 0}~.\\\\2.~\boxed{\bf b_n=b_1\cdot q^{n-1},~\forall~n\in\mathbb{N^*},~n\geq 2,~q\neq 0}~.\\\\3.~\boxed{\bf S_n=\frac{b_1(q^n-1)}{q-1},~\forall~n\in\mathbb{N^*},~n\geq 2,~q\neq 1}~.\\-----------------------------$

$\displaystyle\bf\\\boxed{\bf 1}~.~a_1=4,~r=6,~atunci~a_2=a_1+r=\boxed{\bf 10}~.\\\\\boxed{\bf 2}~.~a_1=4,~r=6,~atunci~a_5=a_1+(n-1)r=4+4\cdot6=4+24=\boxed{\bf28}~.\\\\\boxed{\bf 3}~.~a_1=4,~r=6~stim~a_5=28~de~la~problema~2,~atunci~:~\\\\S_5=\frac{(a_1+a_5)5}{2} =\frac{(4+28)5}{2}=80~.\\\boxed{\bf 4}~.~b_1=5,~r=2,~b_2=b_1r=10.\\\\\boxed{\bf 5}~.~b_1=5,~r=2,~b_5=b_1\cdot r^{n-1}=5\cdot 2^4=80.\\\\\boxed{\bf 6}~.~b_1=5,~r=2,~b_5=80,~S_5=\frac{b_1(q^n-1)}{q-1}=\frac{5\cdot(2^5-1)}{2-1} =155.$

$\displaystyle\bf\\\boxed{\bf 7}~.~a_n=\frac{n-1}{n-2},~a_8=\frac{8-1}{8-2}=\frac{7}{6}~.\\\\\boxed{\bf 8}~.~a_n=(-1)^n+6,~a_12=(-1)^{12}+6=1+6=7.$