SOLUTION: Given the arithmetic series : 12+18+24+....+2880 Calculate the sum of all the terms of the series tht are not visible by 4 6x+12;2x+4;andx-7 are the first three terms of a geo

Algebra ->  Sequences-and-series -> SOLUTION: Given the arithmetic series : 12+18+24+....+2880 Calculate the sum of all the terms of the series tht are not visible by 4 6x+12;2x+4;andx-7 are the first three terms of a geo      Log On


   

Question 1083503: Given the arithmetic series : 12+18+24+....+2880
Calculate the sum of all the terms of the series tht are not visible by 4
6x+12;2x+4;andx-7 are the first three terms of a geometric sequence
Solve for x

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The common difference is 6, and A\all terms are divisible by 6,
from 6%2A2=12=4%2A3 to 6%2A480=2880=4%2A180%2A6 .
First and last terms 12 and 2880 are divisible by 4.
However, with a common difference of 6=3%2A2 ,
every other term will be not divisible by 4.
Those terms from an arithmetic series with common difference 6%2A2=12 ,
from a%5B1%5D=6%2A3=18 to a%5Bn%5D=6%2A479=2874 .
The difference between first and last term is
a%5Bn%5D-a%5B1%5D=12%28n-1%29=6%2A479-6%2A3=6%28479-3%29=6%2A476=6%2A2%2A238=12%2A238 .
So, n-1=238 ---> n=239
The sum of those n terms is
n%28a%5B1%5D%2Ba%5Bn%5D%29%2F2=239%2818%2B2874%29%2F2=239%2A2892%2F2=highlight%28345594%29

The common ratio of that geometric sequence is
%28x-7%29%2F%282x%2B4%29=%282x%2B4%29%2F%286x%2B12%29
Solving:
%28x-7%29%2F%282x%2B4%29=%282x%2B4%29%2F%286x%2B12%29
%28x-7%29%2F%282%28x%2B2%29%29=%282x%2B4%29%2F%286%28x%2B2%29%29
Denominators require x%2B2%3C%3E0%29 <--> x%3C%3E-2
Multiplying both sides times 12%28x%2B2%29 , we gw=et
6%28x-7%29=2%282x%2B4%29
6x-42%29=4x%2B8
6x-4x=42%2B8
2x=50
highlight%28x=25%29