Гкометрическая прогрессия b₁=5 q=3 Sn=200 n-?. Created by Besecond. algebra-ru

Гкометрическая прогрессия b₁=5 q=3 Sn=200 n-?...

July 2022 1 9 Report

Гкометрическая прогрессия b₁=5 q=3 Sn=200 n-?

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Apofeoz

$S_n = \frac{b_1(1-q^n)}{(1-q)};$

$S_n(1-q)=b_1(1-q^n);$

$S_n-S_nq=b_1-b_1q^n;$

$S_n-S_nq-b_1=-b1q^n;$

$q^n=\frac {S_n-S_nq-b_1}{-b_1};$

$3^n=\frac{(200-200*3-5)}{-5};$

$3^n=81;$

$n = 4;$

Ответ: $n = 4.$

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