The first term of a geometric series is 2, the nth term is 486 and the sum of the n terms is 728. find r and n.
We have two things we can work with, here........first......the sum of the series
728 = 2 [1 - r^n] / [ 1 - r]
364 = [ 1 - r^n ] / [1 -r]
364 -364 r = 1 - r^n
364r - 364 = r^n - 1
364r - 363 = r^n (1)
And, second.......the nth term
486 = 2(r)^(n-1)
243 = r^(n-1)
243 = r^n / r
243r = r^n (2)
So....equating (1) with (2), we have
364r - 363 = 243r simplify
121r - 363 = 0
121r = 363 divide both sides by 121
r = 3
And using (2)
243(3) = 3^n
729 = 3^n
And n = 6
Sorry.....I get the same result as Melody........check
Sum = 2[ 1 - 3^6] / [ 1 -3] = 728
And the 6th - and last - term is given by
2*3^(6 -1) =
2*3^5 =
2* 243 =
486
I'm not sure of the answers generated by WA........for instance.........their calcuation for "n" results in a decimal approximation......it should be an integer!!!!......I'd go with the solutions Melody and I provided
