precalc help please, giving medals and becoming fan!!! 1.) Find the first six terms of the sequence. a1 = -8, an = 5 • an-1 -8, -40, -200, -1000, -5000, -25,000 -8, -40, -35, -30, -25, -20 0, 5, -40, -35, -30, -25 -40, -200, -1000, -5000, -25,000, -125,000 2.)Find an equation for the nth term of the arithmetic sequence. -3, -5, -7, -9, ... an = -3 - 2 an = -3 + -2(n) an = -3 + -2(n + 1) an = -3 + -2(n - 1)

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precalc help please, giving medals and becoming fan!!! 1.) Find the first six terms of the sequence. a1 = -8, an = 5 • an-1 -8, -40, -200, -1000, -5000, -25,000 -8, -40, -35, -30, -25, -20 0, 5, -40, -35, -30, -25 -40, -200, -1000, -5000, -25,000, -125,000 2.)Find an equation for the nth term of the arithmetic sequence. -3, -5, -7, -9, ... an = -3 - 2 an = -3 + -2(n) an = -3 + -2(n + 1) an = -3 + -2(n - 1)

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For the first one let's get you started. $$ \Large {a_1=-8 \\ a_2 = 5\cdot a_1= 5 \cdot (\text{-}8) }$$
similarly, to get a3, plug in n=3 in \(\Large a_n = 5 a_{n-1} \\ \Large a_3 = 5a_2 = ... ?\) and so on..
yu are still stuck?

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for the second one -3, -5, -7,- 9.... do you see that we are adding 2 each step
-2 i meant*
so a1=-3 a2=-3-2=-5 a3=-3-2-2=-7 a4=-3-2-2-2=-9 . . . . an=-3-2-2-2-2-2-2............-2 (-2 (n-1) times)=3-2(n-1)
correction an=-3-2(n-1)

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