The 3rd and the 6th terms of a G.P. are 108 and -32 respectively. Find the sum of the 1st 7 terms

Sure, let's solve this step by step.

1. In a geometric progression (G.P.), the nth term is given by ar^(n-1), where a is the first term and r is the common ratio.

2. Given that the 3rd term is 108, we can write the equation: ar^2 = 108 ...(equation 1)

3. Given that the 6th term is -32, we can write the equation: ar^5 = -32 ...(equation 2)

4. Dividing equation 2 by equation 1, we get r^3 = -32/108 = -8/27. Since r is the cube root of -8/27, r = -2/3.

5. Substituting r = -2/3 in equation 1, we get a(-2/3)^2 = 108, which simplifies to a = 108 / (4/9) = 243.

6. The sum S of the first n terms of a G.P. is given by S = a(r^n - 1) / (r - 1) if r ≠ 1.

7. Substituting a = 243, r = -2/3, and n = 7 in the above formula, we get S = 243((-2/3)^7 - 1) / (-2/3 - 1) = -243.

So, the sum of the first 7 terms of the G.P. is -243.