Question

in progresia geometrica(bn)n se cunoaste ca b3=4 sib9=256.Sa se calculeze b1si s10.​

Answer 1

Răspuns:

Explicație pas cu pas:

b₃ = 4 ;  b₉ = 256

b₃ = b₁·q²  = 4

b₉ = b₁·q⁸  = 256

(b₁·q⁸ ):(b₁·q² ) =256:4  = 64 =>

q⁶ = 64 = > q₁ = 2  ; q₂ = -2

q₁ = 2 ; q₂ = -2  => b₁ = b₃ : q² = 4:4 => b₁= 1

q₁ = 2 =>

S₁₀ = b₁+b₂+b₃+.....+b₁₀

b1 = 1 ; b₂ = 2 ; b₃ = 4 ; b₄ = 8 ......b₁₀ = 512 =>

S₁₀ = 1+2+4+8+....+512 = 1+ (2+512)·[(512-2):2+1] :2 = >

S₁₀ = 1 + 514·256:2 = 1+ 514·128 = 65793

q₂ = -2  =>

S₁₀ = 1-2+4-8+......+256-512 = 1+4+16+64+256 - 2-8-32-128-512 =>

S₁₀ = -1-4-16-64-256 = -341