Question

attach file need help,Please! show your work ,Ty in advance

Answer 1

For the first n terms of a geometric series, the sum is
\[a \frac{1-r^n}{1-r}\]The initial term is a=-1. The common ratio is r=-3. You're looking for a value of n where the sum is 182.
\[182=(-1)\frac{1-(-3)^n}{1-(-3)}\]
Now it's just a bit of algebra:
\[182=(-1)\frac{1-(-3)^n}{1+3}\]\[182=-\frac{1}{4}(1-(-3)^n)\]\[-728=1-(-3)^n\]\[-729=-(-3)^n\]\[729=(-3)^n\]
Using a little thinking, we know that n needs to be an even power in order for the result to be a POSITIVE 729. We can therefore disregard the negative in front of the 3.
Take the natural log of both sides.
\[\ln(729)=\ln((3)^n)\]Use the properties of exponents.\[\ln(729)=n(\ln(3))\]\[n=\frac{\ln(729)}{\ln(3)}=6\]

Answer 2

ty, ur smart

Answer 3

No worries :)