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georgejafacules
 3 years ago
Suppose the first term of a geometric sequence is multiplied by a nonzero constant, c. What happens to the following terms in the sequence? What happens to the sum of this geometric sequence? (This question has one right answer.) Give an example of a geometric sequence to illustrate your reasoning. (Many answers are possible.)
georgejafacules
 3 years ago
Suppose the first term of a geometric sequence is multiplied by a nonzero constant, c. What happens to the following terms in the sequence? What happens to the sum of this geometric sequence? (This question has one right answer.) Give an example of a geometric sequence to illustrate your reasoning. (Many answers are possible.)

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Jnlucero
 3 years ago
Best ResponseYou've already chosen the best response.1the next terms in the sequence will be multiplied by c, but the common ratio is still the same... so let m,n,o,p,.... be the gometric sequence with common ratio r. thus the sequence cm,cn,co,cp,...... will also have the common ratio r. so the sum will be\[S_{n}=\frac{ a_1r^na_1 }{ r1 } = \frac{ cmr^ncm }{ r1 }=c\frac{ mr^nm }{ r1 }\]. take note that the first factor is c and the second is the sum of the sequence m,n,o,p,.....

Jnlucero
 3 years ago
Best ResponseYou've already chosen the best response.1so the sum of the sequence m,n,o,p,..... when multiplied by c gives the sum of the sequence cm,cn,co,cp,......

Jnlucero
 3 years ago
Best ResponseYou've already chosen the best response.1example..... 1,3,9,27, here, r=3. let the multiplier be 4. then the sequence 4, 12,36,108, has still the common ratio 3. for 1,3,9,27, \[S=\frac{ 1(3)^41 }{ 31 }=80/2=40\] for 4,12,36,108\[S=\frac{ 4(3)^4 4 }{ 31 }=320/2 = 160\] thus it was shown
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