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# Calculate, to four decimal places, the first ten terms of the sequence and use them to plot the graph of the sequence by hand. Does the sequence appear to have a limit? If so, calculate it. If not, explain why.$a_n = \frac {3n}{1 + 6n}$

## $\frac{1}{2}$

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for in equals one plug in one. Here we get three over seven, which is approximately zero point for two eight six okay and equals two. We plug in the value we get six over thirteen. Um, six over thirteen is approximately zero point for six one five. Ah, let me just list out all of the values here. So the first ten terms that were going to get zero point for two eight six, that's the first one That's for n equals one. Then we have zero point for six one five. Then we have zero point for seven, three seven and then we'LL have zero point for a hello then we'll have zero point for eight three nine Then we'LL have zero point for eight, six five Okay? And there's going to be, you know, somewhere terms that we would want to include there as well if we I really did all ten of them I'll start drawing this graph So there's one two, three for five six love in along this axis And then the eyes of an over here. So zero point one your point two point three their point for zero point five Okay, so we know the first term is zero point four two eight six. That's when in his one. So that's going to be around here. Next term, zero point four six one five. That's for n equals two sit n equals two for six. So about halfway between point for one point five for in equals three zero point four seven three seven. So that's going to be pretty close to where the other one wass Okay, And then the fourth one point for eight. So that's still going to be pretty close to the other one. Okay. And then you can see the pattern here terms if we just connect the dots. Looks like we're getting pretty close. Looks like we're flattening out here. It looks like we're just getting closer and closer to zero point five. So it looks Yeah, it looks like we converge because it looks like we've converged because it looks like we're flattening out. So if this does flat now and we do happen to have a horizontal ass in tow at zero point five, then that would mean that we converged to zero point five. So we do appear to have a limit because we do appear to have a horizontal asked meto. So to figure out the limit, we would need to. It's right. Lim is an approaches infinity of three in over one plus six and and then this limit shouldn't be too bad to figure out. We can just divide the top on the bottom by and as in goes to infinity one over and goes to zero. So this is this goes to three over six, which is one half, so that is indeed the limit.

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