What is the sum of the geometric series in which a1 = 7, r = 3, and an = 1,701? hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r ≠ 1, where a1 is the first term and r is the common ratio

My best interpretation of this is that you're given a geometric progression consisting of terms, the first of which is and the last of which is . (So we don't actually know right away how many terms there are.) The common ratio between terms is . You want to find the sum of all terms.

In a geometric progression, the -th term is determined by the previous term according to

Starting with , we find

and so on. The general pattern for the -th term is then

The last term in the sequence is , so

The sum of these terms is given by

Notice that

With and , we get a sum of

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.