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Find all possible third terms of a five-term geometric progression of real numbers, the sum of whose term is 16 times the sum of their reciprocals.

 Dec 10, 2019
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a=1; b=(a);c=(a*2);d=(a*4);e=(a*8);f=(a*16); g=if(b+c+d+e+f==16*(1/b+1/c+1/d+1/e+1/f), goto loop, goto next);loop:printa,", ",;next: a++;if(a<=1000000, goto1, 0):


The sequence is:


Sum = 1, 2, 4, 8, 16 = 31
Sum of their reciprocals:
1/1 + 1/2 +1/4 + 1/8 + 1/16 =1.9375


The sum of the sequence =16 x sum of their reciprocal =1.9375 = 31. And that is the end!!.
P.S. I checked all the numbers up to 1,000,000 and this is the only sequence of its kind.

 Dec 10, 2019
edited by Guest  Dec 10, 2019

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