In a geometric series first term is a =3 and n=3 and sum of terms =63

Question

now how to evaluate the value of common ratio r , and last term we know that first we have to evaluate the value of r , and we also know that there are two formulas where r can be greater than or lesser than 1. please tell me how to solve this type of question where we gonna find r, and r can be < or > than 1 what should we do when this type of problem creates.

Answer 1

Given that a = 3, n = 3 and sum of terms is 63

So, the geometric series is a + ar + ar² = 63

substitute a = 3 in the geometric series.

3 + 3r + 3r² = 63

Take out common term 3.

3(1 + r + r²) = 63

Divide each side by 3.

r² + r + 1 = 21

Subtract 21 from each side.

r² + r - 20 = 0

Now solve the factor method.

r² + 5r - 4r - 20 = 0

r(r + 5) - 4(r + 5) = 0

Take out common factors.

(r - 4)(r + 5) = 0

r - 4 = 0 or r + 5 = 0

Take r - 4 = 0

Add 4 to each side

r = 4

And  r + 5 = 0

Subtract 5 from each side.

r = -5.

There fore r = 4 or -5...

r = -5 or 4...both would yield divergent infinite