[tex]S_{10}=\frac{(a_{1}+a_{10})10}{2}=420[/tex]Răspuns:
Explicație pas cu pas:
1. ab,b²,c² progresie aritmetica ⇒2b²=ab+c²
c²=2b²-ab
Trebuie sa demonstram ca b,c,2b-a sunt intr-o progresie geometrica, adica
c²=b(2b-a)
c²=2b²-ab adevarat
2. a₁=33
S₁₀=420
[tex]S_{10}=\frac{(a_{1}+a_{10})\cdot 10}{2}=420[/tex]
a₁+a₁₀=420:5
a₁+a₁₀=84
33+a₁₀=84
a₁₀=51
a₁₀=a₁+9r
51=33+9r
9r=18
r=2
3. a₁+a₃=6
a₄=7
a₁+a₁+2r=6
a₁+3r=7 a₁=7-3r
14-6r+2r=6
8=4r
r=2
4. d-a=7
c-b=2
b=a×q
c=a×q²
d=a×q³
aq³-a=7 a(q³-1)=7
aq²-aq=2 a(q²-q)=2
impartim cele doua ecuatii
[tex]\frac{q^{3}-1 }{q(q-1)} =\frac{7}{2} \\\frac{(q-1)(q^{2}+q+1) }{q(q-1)} =\frac{7}{2}\\\frac{(q^{2}+q+1) }{q} =\frac{7}{2}[/tex]
[tex]\frac{q^{3}-1 }{q(q-1)} =\frac{7}{2} \\\frac{(q-1)(q^{2}+q+1) }{q(q-1)} =\frac{7}{2}\\\frac{(q^{2}+q+1) }{q} =\frac{7}{2}\\[/tex]
2q²+2q+2=7q
2q²-5q+2=0
Δ=25-16=9
q=2
q=[tex]\frac{1}{2}[/tex]
