Впрогрессии найдите b1 и n
q=-1\3; bn=-1\3; sn=182\3

b₁=81; n=6

Объяснение:

Sₙ=(b₁-bₙ·q)/(1-q)

182/3=(b₁-(-1/3)·(-1/3))/(1-(-1/3))

182/3=(b₁ -1/9)/(4/3)

182/3=(b₁ -1/9)·3/4

4·182=9·(b₁ -1/9)

728=9b₁ -9/9

9b₁=728+1

b₁=729/9=81

bₙ=b₁·qⁿ⁻¹

-1/3=81·(-1/3)ⁿ⁻¹

(-1/3)ⁿ⁻¹=(-1/3)/9²

(-1/3)ⁿ⁻¹=(-1/3)/(3²)²

(-1/3)ⁿ⁻¹=(-1/(3·3⁴)

(-1/3)ⁿ⁻¹=(-1/3)⁵

n-1=5

n=5+1=6