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. So, r = -3/3 = -1.

5. Substituting r = -1 in Equation 1, we get a*(-1)^2 = 108, so a = 108.

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

7. Substituting a = 108, r = -1, and n = 7 in the above formula, we get S = 108*(1 - (-1)^7) / (1 - (-1)) = 108*(1 - (-1)) / 2 = 108*2 / 2 = 108.

So, the sum of the first 7 terms of the G.P. is 108.