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

$S_7-S_7r=S_7(1-r)=a_1-a_1r^7\implies S_7=a_1\dfrac{1-r^7}{1-r}$

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

$S_7=(-11)\dfrac{1-(-4)^7}{1-(-4)}=144,188$