kmeds16 3 years ago Given a geometric sequence whose sum of the first 10 terms is 4 and whose sum from the 11th to the 30th term is 48, find the sum from the 31st to the 60th term.

1. cwrw238

Sum of 10 = a * (r^4 - 1) -------- = 4 r- 1

2. kmeds16

@cwrw238 why is it r^4? I thought it's r^10.

3. hitten101

n is 10 not 4

4. amistre64

hmm, given is: $S_n=\frac{1-r^n}{1-r}$ $S_{10}=4=\frac{1-r^{10}}{1-r}$ $S_{30-10}=48=\frac{1-r^{20}}{1-r}$

5. kmeds16

@hitten101 yes yes :)

6. shubhamsrg

you have been given the sum upto first 10 terms =4 you have also been given the sum upto first 30 terms = 4 + 48 =52 and you have 2 eqns with 2 variables ->solve for a and r now calculate sum for first 60 terms from that subtract sum of first 30 terms.. this should help..

7. hitten101

|dw:1351514867224:dw|

8. kmeds16

a(r^10 - 1) / r - 1 = 4 a(r^30 - 1) / r - 1 = 52 ? @shubhamsrg like this?

9. hitten101

solve for a and r.. then find the sum of 60 terms subtract sum of 30 terms from sum of 60 terms

10. shubhamsrg

@kmeds16 yep @hitten101 mistake in your formulla in the denominator..

11. kmeds16

I got, r^10 = 3. this is confusing :/ 10th root of 3?!

12. cwrw238

my mistake r^10 not r^4

13. shubhamsrg

how'd you get that? o.O

14. shubhamsrg

ahh k..got it

15. hitten101

@kmeds16 @shubhamsrg yes no exponent in the denominator.. you are right

16. kmeds16

second equation divided by first equation. hehehehehe

17. shubhamsrg

your main aim is not to find r,, your main aim is to find sum.. leave it as r^10 = 3

18. kmeds16

find the sum of S60 and subtract 52, right?

19. shubhamsrg

seems likely..

20. shubhamsrg

do this,,this might simplify.. substitue r^10 =3 whereever you can leave r-1 as it is.. you can see a/(r-1) = 4/(r^10 -1) in calculation for sum of 60 terms ,make use of this eqn,, no need to find a.. :)

21. shubhamsrg

r^60 we all can find.. hmm.. hope that helped..

22. kmeds16

solving...hehehe ahm, thanks for the idea..