Question

What is the sum of a 7-term geometric series if the first term is -6, the last term is -24,576, and the common ratio is 4?

Answer 1

Answer:

Hence, the sum of a 7-term geometric series is:

-32766.

Step-by-step explanation:

We have to find the sum of a 7-term geometric series (i.e. n=7) if the first term(a) is -6, the last term is -24,576, and the common ratio(r) is 4.

We know that the sum of the 7-term geometric series is given as:

$S_n=a\times (\dfrac{r^n-1}{r-1})$

On putting the value of a,n and r in the given formula we have:

$S_7=(-6)\times (\dfrac{4^7-1}{4-1})\\\\\\S_7=-32766$

Hence, the sum of a 7-term geometric series is:

-32766.

Answer 2

Answer:  The required sum of the given geometric series is - 32766.

Step-by-step explanation:  We are given to find the sum of a 7-term geometric series if the first term is -6, the last term is -24,576 and the common ratio is 4.

We know that,

if 'a' is the first term and 'r' is the common ratio of a geometric series, then its sum up to n terms is given by

$S_n=\dfrac{a(1-r^n)}{1-r},~r<1,~~~~~\textup{or}~~~~~S_n=\dfrac{a(r^n-1)}{r-1},~r>1.$

In the given geometric series,

first term, a = -6  and  common ratio, r = 4.

Since r = 4 > 1, so the sum up to 7 terms is

$S_7\\\\\\=\dfrac{a(r^7-1)}{r-1}\\\\\\=\dfrac{-6(4^7-1)}{4-1}\\\\\\=\dfrac{-6(16384-1)}{3}\\\\\\=-2\times 16383\\\\=-32766.$

Thus, the required sum of the given geometric series is - 32766.