The sum of the first 3 terms of a geometric series is 37/8. The sum of the first 6 term is 3367/512. Find the first term and the common ratio?

Let the first term of the geometric series be a and the common ratio be r.

The sum of the first 3 terms is given by:
a + ar + ar^2 = 37/8

The sum of the first 6 terms is given by:
a + ar + ar^2 + ar^3 + ar^4 + ar^5 = 3367/512

We can rewrite the sum of the first 6 terms as:
a(1 + r + r^2 + r^3 + r^4 + r^5) = 3367/512

Now, we can use the formula for the sum of a geometric series to simplify the expression:
a(1 - r^6)/(1 - r) = 3367/512

We also know that the sum of the first 3 terms is 37/8, so we can substitute this into the first equation:
a(1 + r + r^2) = 37/8

Now we have a system of two equations:
1) a(1 + r + r^2) = 37/8
2) a(1 - r^6)/(1 - r) = 3367/512

We can solve this system of equations to find the values of a and r.

Let the first term be a and the common ratio be r.
Then, the sum of the first 3 terms
= a + ar + ar^2
= a (1 + r + r^2)
= 37/8 ….(i)

And, the sum of the first 6 terms
= a + ar + ar^2 + ar^3 + ar^4 + ar^5
=a (1 + r + r^2 + r^3 + r^4 + r^5)
= 3367/512 ….(ii)

Dividing (ii) by (i), we get
(1 + r + r^2 + r^3 + r^4 + r^5)/(1 + r + r^2) = 3367/512 - 8/37
= (1 + r)^3/(1 + r + r^2) = 64/27
= (1 + r)^3 = 64/27 - (1 + r + r^2)
= 1 + 3r + 3r^2 + r^3 = 64/27 + 64/27 - r + 64/27 - r^2
= 1 + 3r + 3r^2 + r^3 = 64/27 + 64/27 - r + 64/27 - r^2
= r^3 + 3r^2 + 3r + 1 = 64/27 + 64/27 - r + 64/27 - r^2

Comparing the coefficients of like powers, we get
a = 1, r^2 + 3r + 1 = 64/27, r = -8/9

Therefore, the first term is 1 and the common ratio is -8/9.