Question

what is the sum of a 7-term geometric series if the first term is negative 11 the last turn is -45056 and the common ratio is -4

Answer 1

For a geometric sequence with common ratio
`r`, we have


`a_n=a_(n-1)r`


that is, the
`n`-th term in the sequence is the product of the previous term and the common ratio
`r`. So


`a_n=a_(n-1)r=a_(n-2)r^2=\cdots=a_1r^(n-1)`

Then the sum of the first 7 terms is


`S_7=\displaystyle\sum_(n=1)^7a_n=a_1+a_2+\cdots+a_7`

`\implies S_7=a_1+a_1r+\cdots+a_1r^6`

Notice that


`S_7r=a_1r+a_1r^2+\cdots+a_1r^7`

so we can subtract this modified sum from
`S_7` to get


`S_7-S_7r=S_7(1-r)=a_1-a_1r^7\implies S_7=a_1(1-r^7)/(1-r)`

We're told that
`a_1=-11` and
`r=-4`, so the sum of the first 7 terms is


`S_7=(-11)(1-(-4)^7)/(1-(-4))=144,188`