Question

85 . найдите шестой и n-ый член прогрессии: а)48; 12; б) дробь 64 деленнач на 9; дробь -32 деленное на 3; в) -0.001; -0.01; г) -100; 10;

Answer 1

А) 48, 12, ,,,   
d = 12 / 48 = 1/4
b6 = b1 * d^5 = 48 * (1/4)^5 = 48/1024 = 6/128=3/64
bn = b1 * (1/4)^(n-1)$b_n = b_1* (\frac{1}{4})^{n-1}$
б) $\frac{64}{9}; - \frac{32}{3} \\ d=- \frac{32}{3}: \frac{64}{9}=- \frac{32}{3}* \frac{9}{64}=- \frac{3}{2} \\ b_6= \frac{64}{9}*(- \frac{3}{2})^5= -\frac{64}{9}* \frac{3^5}{2^5}= -\frac{64}{9}* \frac{243}{32}= -54 \\ b_n=b_1*(- \frac{3}{2})^{n-1}$
в)$-0.001; -0.01 \\ d=-0.01:(-0.001)=10 \\ b_6=(-0.001)*10^5=-100 \\ b_n=(-0.001)*10^{n-1}$
г)$-100, 10 \\ d=10:(-100)=-0.1 \\ b_6=-100*(-0.1)^5=-100*(-0.00001)=0.001 \\ b_n=-100*(-0. 1)^{n-1}$

Answer 2

А) 48, 12,...

b1 = 48, q = 1/4

b6 = b1*q^5 = 48/4^5 = 3/64

bn = b1*q^(n-1) = 48/4^(n-1) = 3*(4^(3-n))

б) 64/9, -32/3...

b1 = 64/9, q = - 3/2

b6 = b1*q^5 = - 64*243/(9*32) = -54

bn = b1*q^(n-1) = (64/9)*(-3/2)^(n-1)

в) -0,001; -0,01...

b1 = -0,001; q = 10

b6 = b1*q^5 = -0,001*10^5 = -100

bn = b1*q^(n-1)= -0,001* 10^(n-1) = -10^(n-4)

г) -100, 10

b1= -100; q = -0,1

b6 = b1*q^5 = 100 *(-10)^(-5) = -0,001

bn = b1*q^(n-1) = 100*(-0,1)^(n-1)