Срочно!!! ГЕОМЕТРИЧЕСКАЯ ПРОГРЕССИЯ b1=3 q=1/3 S=121/27 НАЙТИ ЧИСЛО ЧЛЕНОВ ПРОГРЕССИИ. Created by kristinamaltseva1. algebra-ru

27 НАЙТИ ЧИСЛО ЧЛЕНОВ ПРОГРЕССИИ...

АнжелаКотова $Sn= \frac{b_{1}(q^n-1)}{q-1} \\ \\ 
 \frac{121}{27}= \frac{3(( \frac{1}{3} )^n-1)}{ \frac{1}{3}-1 } \\ \\ 
 \frac{121}{3^3}= \frac{3(( \frac{1}{3} )^n-1)}{- \frac{2}{3} } \\ \\ 
 \frac{121}{3^3}=- \frac{9}{2}(( \frac{1}{3})^n-1) \\ \\ 
 \frac{121}{3^3}*(- \frac{2}{9} )= \frac{1}{3^n}-1 \\ \\ 
- \frac{121*2}{3^3*3^2}= \frac{1}{3^n}-1 \\ \\ 
 -\frac{242}{3^5}+1= \frac{1}{3^n} \\ \\ 
 \frac{3^5-242}{3^5}= \frac{1}{3^n} \\ \\ 
 \frac{243-242}{3^5}= \frac{1}{3^n} \\ \\ 
$
$\frac{1}{3^5}= \frac{1}{3^n} \\ \\ 
3^5=3^n \\ 
n=5$
Ответ: 5 членов прогрессии.

m11m

Verified answer

b2=b1q
b1* b1q=27
b1²q=27
b1²=27/q

b3=b1*q²
b4=b1*q³
(b1)²q*5=1/3
27/q*q*5=1/3
27q*4=1/3
q*4=1/3:27=1/81
q=1/3

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