SOLUTION: The first tem of a geometric progression is 7, its last term is 448 and its sum is 889. Find the common ratio

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Question 732891: The first tem of a geometric progression is 7, its last term is 448 and its sum is 889. Find the common ratio
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
THE LUCKY GUESS APPROACH:
If we call the number of terms n, and the common ratio r, the last term is
b%5Bn%5D=7%2Ar%5E%28n-1%29=448 --> r%5E%28n-1%29=448%2F7 --> r%5E%28n-1%29=64 -->
r%5E%28n-1%29=2%5E6
If we are lucky r=2 and n-1=6 --> n=7.
Those numbers would certainly agree with the first term being 7 and the last term being 448.
Would the sum of those 7 terms be 889?
We can figure out the sum using some formula from the book, or we can calculate terms number 2, 3, 4, 5, and 6 and then add all 7 terms.
Either way, we would find out that the sum of the first 7 terms of a geometric progression with ratio 2 and first term 7 is indeed 889.
The approach worked because the problem was design to have nice small numbers as answers for r and n.

ANOTHER APPROACH:
The sum of n terms, from b%5B1%5D=7 to b%5Bn%5D=448 is 889
The sum of the first n-1 terms, from b%5B1%5D=7 to b%5Bn-1%5D is
S=b%5B1%5D%2Bb%5B2%5D+ ... +b%5Bn-2%5D%2Bb%5Bn-1%5D=889-448=441
If we call the common ratio r,
S%2Ar=b%5B1%5D%2Ar%2Bb%5B2%5D%2Ar+... +b%5Bn-2%5D%2Ar%2Bb%5Bn-1%5D%2Ar=b%5B2%5D%2Bb%5B3%5D+ ... +b%5Bn-1%5D%2Bb%5Bn%5D
because b%5B1%5D%2Ar=b%5B2%5D , b%5B2%5D%2Ar=b%5B3%5D , and so on until b%5Bn-2%5D%2Ar=b%5Bn-1%5D , and b%5Bn-2%5D%2Ar=b%5Bn-1%5D
S%2Ar-S=b%5Bn%5D-b%5B1%5D --> S%2A%28r-1%29=b%5Bn%5D-b%5B1%5D --> r-1=%28b%5Bn%5D-b%5B1%5D%29%2FS --> r=1%2B%28b%5Bn%5D-b%5B1%5D%29%2FS
Since we know b%5Bn%5D , b%5B1%5D, and S we can calculate
r=1%2B%28448-7%29%2F441 --> r=1%2B441%2F441 --> r=1%2B1 --> highlight%28r=2%29