Question

Am nevoie de ajutor la aceste progresii va rog!​

Answer 1

$S_{10}=\frac{(a_{1}+a_{10})10}{2}=420$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

$S_{10}=\frac{(a_{1}+a_{10})\cdot 10}{2}=420$

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

$\frac{q^{3}-1 }{q(q-1)} =\frac{7}{2} \\\frac{(q-1)(q^{2}+q+1) }{q(q-1)} =\frac{7}{2}\\\frac{(q^{2}+q+1) }{q} =\frac{7}{2}$

$\frac{q^{3}-1 }{q(q-1)} =\frac{7}{2} \\\frac{(q-1)(q^{2}+q+1) }{q(q-1)} =\frac{7}{2}\\\frac{(q^{2}+q+1) }{q} =\frac{7}{2}$

2q²+2q+2=7q

2q²-5q+2=0

Δ=25-16=9

q=2

q=$\frac{1}{2}$