Question

Check these please?

Answer 1

@Michele_Laino

Answer 2

from the general theory, we can write this:
\[\frac{{140}}{{35}} = \frac{1}{{{q^2}}}\]
where \(q\) is the constant of the sequence. Please solve for \(q\)

Answer 3

How do you pronounce your name? @Michele_Laino

Answer 4

q=1/2,-1/2

Answer 5

my name is equivalent to "Michael"

Answer 6

ok! the possible values of \(q\) are correct!

Answer 7

Oh good! That's what I have been calling you the whole time! x :)

Answer 8

Sweet, so what now?

Answer 9

now, we have to compute the first term of the sequence. In order to do that, we can write this:
\[35 = {a_1}{q^6} = \frac{{{a_1}}}{{64}}\]
please solve for \(a_1\)

Answer 10

Wait, Michele, how did you get that?

Answer 11

since, from the general theory, we have:
\[{a_7} = {a_1}{q^{\left( {7 - 1} \right)}}\]

Answer 12

I got the first and second one! Hey Michele how about the 3rd one? aye?

Answer 13

Oh, duh. Sorry, brain fart :P

Answer 14

I think that we have to apply this formula:
\[{S_{10}} = {a_1}\frac{{1 - {q^{10}}}}{{1 - q}} = 1 \cdot \frac{{1 - {4^{10}}}}{{1 - 4}} = ...?\]
which comes from the general theory about geometric sequences

Answer 15

349525 awesome thinking!

Answer 16

that's right!