Question

Se considera progresia aritmetica (an) n>1; in care a15 =29 r=2. Sa se calculeze; a1; a2; a6; a10.

Answer 1

$a_{15} = 29$
$r =2$

$a_{15} = a_1 + (15-1)*2 = a_1 + 14*2 = a_1 + 28 \implies a_1 =1$
$a_6 = a_1 + 5*2 = 1+10=11$
$a_2= a_1 + 2 = 3$
$a_{10} = a_1 + 9*2 = 1+18 = 19$

Answer 2

a₁₅=29, r=2

formula generala: an=a₁+(n-1)×r

⇒a₁₅=a₁+(15-1)×2⇒29=a₁+14×2⇒29-28=a₁

⇒a₁=1

a₂=a₁+(2-1)×r=1+1×2=3

⇒a₂=3

a₁₀=a₁+(10-1)×2=1+9×2=1+18=19

⇒a₁₀=19