Pierwszy wyraz ciągu arytmetycznego ( an ) jest równy: - 3a dziesiąty 24. Wyznacz szósty wyraz tego ciągu.
a1 = -3
a10 = 24
a6 = ?
an = a1 + (n - 1) * r
a10 = a1 + (10 - 1) * r
a10 = a1 + 9r
24 = -3 + 9r
24 + 3 = 9r
27 = 9r
r = 3
a6 = a1 + 5r
a6 = -3 + 5 * 3
a6 = -3 + 15
a6 = 12 --------- odpowiedź
a₁ = -3
a₁₀ = 24
a₁₀ = a₁ + 9r
a₁ + 9r = 24
-3 + 9r = 24
9r = 27
a₆ = a₁ + 5r
a₆ = -3 + 5*3 = -3 + 15 = 12
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a1 = -3
a10 = 24
a6 = ?
an = a1 + (n - 1) * r
a10 = a1 + (10 - 1) * r
a10 = a1 + 9r
24 = -3 + 9r
24 + 3 = 9r
27 = 9r
r = 3
a6 = a1 + 5r
a6 = -3 + 5 * 3
a6 = -3 + 15
a6 = 12 --------- odpowiedź
a₁ = -3
a₁₀ = 24
a₁₀ = a₁ + 9r
a₁ + 9r = 24
a₁ = -3
-3 + 9r = 24
9r = 27
r = 3
a₆ = a₁ + 5r
a₆ = -3 + 5*3 = -3 + 15 = 12
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)