The common ratio of an infinite geometric series is 11/16, and its sum is 76 4/5, how do you find the first four terms of the series?

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Answer 1 · 2017-08-23T00:44:50

The first four terms of the series are:

#24, 33/2, 363/32, 3993/512#

The general term of a geometric series is given by the formula:

#a_n = ar^(n-1)#

where #a# is the initial term and #r# the common ratio.

The sum of the first #N# terms of such a series is:

#s_N = (a(1-r^N))/(1-r)#

If #abs(r) < 1# then #lim_(N->oo) r^N = 0# and the sum of the whole series is:

#s_oo = lim_(N->oo) s_N = a/(1-r)#

In our example, we are told:

#{ (r = 11/16), (s_oo = 76 4/5 = 384/5) :}#

Hence:

#384/5 = s_oo = a/(1-r) = a/(1-11/16) = (16a)/5#

Multiplying both ends by #5/16# we find:

#a = 24#

So the first four terms are:

#24#

#24*11/16 = 33/2#

#33/2*11/16 = 363/32#

#363/32*11/16 = 3993/512#