anonymous one year ago 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

1. anonymous

@Michele_Laino

2. anonymous

@dan815

3. anonymous

@pooja195

4. anonymous

@Preetha

5. amoodarya

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

6. Michele_Laino

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

7. Michele_Laino

where q=-5 and a_1= -11

8. amoodarya

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

9. anonymous

I got -716147.6667

10. Michele_Laino

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

11. anonymous

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

12. anonymous

I got this wrong, right?

13. Michele_Laino

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

14. anonymous

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

15. anonymous

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

16. anonymous

I got it is that right?

17. Michele_Laino

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

18. anonymous

But it's positive

19. Michele_Laino

yes!

20. anonymous

Yes its positive negative times negative

21. anonymous

Thanks! Can you help me with my next question

22. Michele_Laino

ok!

23. anonymous

What is the 7th term of the geometric sequence where a1 = 1,024 and a4 = -16? 1 -0.25 -1 0.25

24. Michele_Laino

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

25. Michele_Laino

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

26. anonymous

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

27. Michele_Laino

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

28. Michele_Laino

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

29. anonymous

Yes

30. anonymous

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

31. Michele_Laino

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

32. anonymous

0.25?

33. Michele_Laino

sorry, I got another result

34. anonymous

I got now -.0625

35. Michele_Laino

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

36. anonymous

Thanks!

37. Michele_Laino

:)