Resposta:
Explicação passo a passo:
primeiro termo =x
razão= 3
soma = 160
montando temos.
x +xr+xr²+xr³ =160
x( 1+r+r²+r³= 160
x(1+1+3+9+27)=160
40x= 160
x= 160/40
x= 4
[tex] > \: resolucao \\ \\ \geqslant \: progressao \: \: geometrica \\ \\ sn = \frac{a1(q {}^{n} - 1) }{q - 1} \\ \\ 160 = \frac{a1(3 {}^{4} - 1) }{3 - 1} \\ \\ 160 = \frac{a1(81 - 1)}{2} \\ \\ 160 = \frac{a1(80)}{2} \\ \\ 320 = 80a1 \\ \\ a1 = \frac{320}{80} \\ \\ a1 = 4 \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \geqslant [/tex]
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Resposta:
Explicação passo a passo:
primeiro termo =x
razão= 3
soma = 160
montando temos.
x +xr+xr²+xr³ =160
x( 1+r+r²+r³= 160
x(1+1+3+9+27)=160
40x= 160
x= 160/40
x= 4
[tex] > \: resolucao \\ \\ \geqslant \: progressao \: \: geometrica \\ \\ sn = \frac{a1(q {}^{n} - 1) }{q - 1} \\ \\ 160 = \frac{a1(3 {}^{4} - 1) }{3 - 1} \\ \\ 160 = \frac{a1(81 - 1)}{2} \\ \\ 160 = \frac{a1(80)}{2} \\ \\ 320 = 80a1 \\ \\ a1 = \frac{320}{80} \\ \\ a1 = 4 \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \geqslant [/tex]