Odpowiedź:
zad 1
an = a₁ + (n - 1)r
{a₁₀ = a₁ + 9r = 4
{a₁₅ = a₁ + 14r = 3
---
{a₁ = 4 - 9r
{4 - 9r + 14r = 3
---
{a₁ = 4 - 9r
{5r = -1 |:5
---
{a₁ = 4 - 9r
{r = -1/5 = -0,2
---
{a₁ = 4 - 9*(-0,2)
{r = -0,2
---
{a₁ = 4 + 1,8
{r = -0,2
---
{a₁ = 5,8
{r = -0,2
a₂₀ = a₁ + 19r
a₂₀ = 5,8 + 19*(-0,2)
a₂₀ = 5,8 - 3,8
a₂₀ = 2
S₂₀ = [(a₁ + a₂₀)*20]/2
S₂₀ = (5,8 + 2)*10
S₂₀ = 78
zad 2
[tex]\cfrac{27^{15}*9^{-25}}{3^{-8}}=\cfrac{(3^{3})^{15}*(3^{2})^{-25}}{3^{-8}}=\cfrac{3^{45}*3^{-50}}{3^{-8}}=\\
\\
=3^{45+(-50)}:3^{-8}=3^{-5-(-8)}=3^{-5+8}=3^{3}=27\\
\\
\\
(a^{n})^{m}=a^{n*m}\\
a^{n}*a^{m}=a^{n+m}\\
a^{n}:a^{m}=a^{n-m}\\
[/tex]
Szczegółowe wyjaśnienie:
