Question

FAN/MEDAL
What is the sum of an 8-term geometric series if the first term is -11, the last term is 859,375, and the common ratio is -5?

-143,231
-36,047
144,177
716,144

Answer 1

@Michele_Laino

Answer 2

@dan815

Answer 3

@pooja195

Answer 4

@Preetha

Answer 5

\[S_8?\\a_1=-11\\a_8=859,375\]
\[s_n=a_1+a_1q+a_1q^2+...+a_1q^{n-1}=\frac{a_1(1-q^n)}{1-q}\\s_8=a_1+a_1q+a_1q^2+...+a_1q^{7}=\frac{a_1(1-q^8)}{1-q}\]

Answer 6

hint:
the sum of the first 8 terms of your geometric sequence is given by the subsequent formula:
\[\Large S = {a_1}\frac{{{q^8} - 1}}{{q - 1}}\]

Answer 7

where q=-5
and a_1= -11

Answer 8

you can write it as
\[a_8=a_1q^7=-11(q^7)=859,375\\q=-5\]
plug into formula

Answer 9

I got -716147.6667

Answer 10

substituting your data into the formula for S, we get:
\[\Large S = {a_1}\frac{{{q^8} - 1}}{{q - 1}} = \left( { - 11} \right) \times \frac{{{{\left( { - 5} \right)}^8} - 1}}{{\left( { - 5 - 1} \right)}} = ...?\]
please continue

Answer 11

-390625-1 (-11) -390626
---------- x ---- = -------- = -65104.33333 x -11 = -716147.6667
-6 1 6

Answer 12

I got this wrong, right?

Answer 13

please keep in mind that:
\[\Large {\left( { - 5} \right)^8} = 390625\]

Answer 14

But then I added to there is a 1- in front

Answer 15

3906254 (-11) 39062
---------- x ---- = -------- = -65104 x -11 = -716144
-6 1 - 6

Answer 16

I got it is that right?

Answer 17

\[{\left( { - 5} \right)^8} = {\left( { - 1 \times 5} \right)^8} = {\left( { - 1} \right)^8}{\left( 5 \right)^8} = + {5^8} = 390625\]
hint:
\[S = {a_1}\frac{{{q^8} - 1}}{{q - 1}} = \left( { - 11} \right) \times \frac{{{{\left( { - 5} \right)}^8} - 1}}{{\left( { - 5 - 1} \right)}} = \left( { - 11} \right) \times \frac{{390625 - 1}}{{\left( { - 5 - 1} \right)}} = ...?\]

Answer 18

But it's positive

Answer 19

yes!

Answer 20

Yes its positive negative times negative

Answer 21

Thanks! Can you help me with my next question

Answer 22

ok!

Answer 23

What is the 7th term of the geometric sequence where a1 = 1,024 and a4 = -16?

1
-0.25
-1
0.25

Answer 24

we have the subsequent formulas:
\[\begin{gathered}
{a_4} = {a_1}{q^3} \hfill \\
{a_7} = {a_1}{q^7} \hfill \\
\end{gathered} \]

Answer 25

so we can re-write the first formula as below:
\[ \Large - 16 = 1024 \times {q^3}\]
please solve for q

Answer 26

-16/1024 = 1024 x q^3/1024
-.015625=q^3
(3)sqrt(-.015625)=(3)sqrt(q^3)
q=.53860

Answer 27

the computation -.015625=q^3 is right, so we have:
\[\large q = \sqrt[3]{{ - 0.015625}} = ...?\]

Answer 28

I got:
q=-1/4
am I right?

Answer 29

Yes

Answer 30

Oh I got it now, I forgot to put the negative on the calculator

Answer 31

ok! Now we have:
\[\Large {a_7} = {a_1}{q^7} = 1024 \times {\left( { - \frac{1}{4}} \right)^7} = ...?\]

Answer 32

0.25?

Answer 33

sorry, I got another result

Answer 34

I got now -.0625

Answer 35

yes!
\[\Large {a_7} = - 0.0625 = - \frac{1}{{16}}\]

Answer 36

Thanks!

Answer 37

:)