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.
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.
First term=2
The ratio =3
The 6th and the last term=486
Sum of G.P =728
+118612 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.
\(a=2 \qquad T_n=486\qquad S_n=728\\~\\ ar^{n-1}=486 \qquad \frac{a(r^n-1)}{r-1}=728\\ 2*r^{n-1}=486 \qquad \frac{2(r^n-1)}{r-1}=728\\ r^{n-1}=243 \qquad \frac{(r^n-1)}{r-1}=364\\ \)
\(\frac{r^n}{r}=243 \qquad \qquad r^n-1=364r-364\\ r^n=243r \qquad\qquad r^n=364r-363\\ 243r=364r-363\\ 363=121r\\ r=3\\~\\ 3^n=243*3\\ 3^n=3^6\\ n=6\\ so\\~\\ n=6\qquad and \qquad r=3\)
CPhill: Can you please look at this Wolfram/Alpha "solution" that Melody got? I don't get it and neither does W/A!!!!!. Thanks.
http://www.wolframalpha.com/input/?i=(+r%5E(n+-+1))+%2F+(r+-+1)%3D364,+r%5E(n-1)%3D243,+solve+for+r,+n
+128707 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
