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Максеня
@Максеня
August 2022
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в геометрической прогрессии n=5,q=1/3,Sn=121.НАйдите первый и n-й член геометрической прогрессии?
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Grammulka
Sn=b1(q^n-1)/(q-1).
b1=Sn*(-2/3)/(-242/243))=121*2/3*243/242=243/3=81.
bn=b5=81*(1/3)^4=81/81=1.
Ответ: b1=81, bn=1.
8 votes
Thanks 32
AlbuRin
N = 5, q = 1/3, S_n = 121. b_1 - ? b_n - ?
S_n = b_1 (q^n - 1)/(q - 1)
121 = b_1 ((1/3)^5 - 1)/(1/3 - 1)
b_1 =121/( (-242/243)/(-2/3) ) = 121/(121/81) = 121*(81/121) = 81
b_n = b_1 * q^(n - 1)
b_5 = b_1 * q^4 = 81*(1/3)^4 = 81*(1/81) = 1
Ответ. b_1 = 81, b_5 = 1
7 votes
Thanks 20
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Answers & Comments
b1=Sn*(-2/3)/(-242/243))=121*2/3*243/242=243/3=81.
bn=b5=81*(1/3)^4=81/81=1.
Ответ: b1=81, bn=1.
S_n = b_1 (q^n - 1)/(q - 1)
121 = b_1 ((1/3)^5 - 1)/(1/3 - 1)
b_1 =121/( (-242/243)/(-2/3) ) = 121/(121/81) = 121*(81/121) = 81
b_n = b_1 * q^(n - 1)
b_5 = b_1 * q^4 = 81*(1/3)^4 = 81*(1/81) = 1
Ответ. b_1 = 81, b_5 = 1