Wyznacz i r ciągu arytmetycznego (an), jeśli
a)
b) wyznacz a ciągu geometrycznego (an)
jeśli a_{1}
DAJE NAJ
а)
{a10 - 2*a7 = -5
{a6 - 2*a4 = 1
{a1+9r-2*(a1+6r) = -5
{a1+5r - 2*(a1+3r) = 1
{-a1-2r = -5 /*(-1){-a1-r = 1
+ {a1+2r=5+ {-a1-r = 1
==========
r = 6
a1 = 5-2r = 5-2*6 = 5-12 = -7
b)
{a1+a3 = 39
{a2+a4 = 26
{a1 + a1*q^2 =39{a1*q^1 + a1*q^3 = 26
{a1* (1+q^2) = 39{a1 * (q + q^3) = 26
{ a1 = 39/(1+q^2){ 39(q + q^3) /(1+q^2) = 26 (*)
(*)
26+26q^2 = 39q + 39q^3
q= 2/3
a1 = 39/(1+q^2) = 39/(9/9+4/9) = 39/ (13/9) = 27
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а)
{a10 - 2*a7 = -5
{a6 - 2*a4 = 1
{a1+9r-2*(a1+6r) = -5
{a1+5r - 2*(a1+3r) = 1
{-a1-2r = -5 /*(-1)
{-a1-r = 1
+ {a1+2r=5
+ {-a1-r = 1
==========
r = 6
a1 = 5-2r = 5-2*6 = 5-12 = -7
b)
{a1+a3 = 39
{a2+a4 = 26
{a1 + a1*q^2 =39
{a1*q^1 + a1*q^3 = 26
{a1* (1+q^2) = 39
{a1 * (q + q^3) = 26
{ a1 = 39/(1+q^2)
{ 39(q + q^3) /(1+q^2) = 26 (*)
(*)
26+26q^2 = 39q + 39q^3
q= 2/3
a1 = 39/(1+q^2) = 39/(9/9+4/9) = 39/ (13/9) = 27