find the sum of the first n turns of a geometric series with a1=3, an=768 and r=-2

Answer 1

From the above help, you will also need to use the formula A_n=A_1r^(n-1) this will give you the n needed to plug into the above formula and find the sum of the nth term in this geometric series.

Answer 2

First let's find out which term 768 is.

a_n = a₁ r^n-1 a_n = 768 a₁ = 3 r=-2

768 = 3 (-2)^n-1 (-2)^n-1 = 256 n-1 = 8 n = 9

Sum Formula S_n = a₁(1-rⁿ)/(1-r) a₁ =3 n=9 r= -2

S_n = 3 (1-(-2)⁹)/ (1-(-2) S_n =3(1-(-512))/3 S_n = 1-(-512) = 513

Answer 3

You just use the formula: geometric sum of n terms= S_n = (a((r^n)-1))/(r-1)

It is self-explanatory you just need to know the respective terms: S_n = sum of n terms and it is not term: (an) it is (a_n) = "a sub n" and plug in (a_1) = "a sub 1" to find respective terms in formula S_n. Make sure when you do this problem you understand what each term means and make sure to write it out properly. Your confusion lies with the misunderstanding of the respective terms or you do not know how to do the Algebra in this problem.