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.

Question

cwrw238 · Answer

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

anonymous · Answer

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

anonymous · Answer

n is 10 not 4

amistre64 · Answer

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

anonymous · Answer

@hitten101 yes yes :)

shubhamsrg · Answer

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..

anonymous · Answer

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