The sum of three consecutive terms of a geometric sequence is 73.5. If the product of these terms is 2744, what are the terms?
56 , 14 and 3.5 56+14+3.5=73.5 56*14*3.5=2744
How did you get that?
this question was asked during one of my mathematical examinations..i still remember it as a matter of fact
WOAH. That is awesome. SO how did you solve for it when you were taking the exam?
Let x be the first element and r be the common ratio x + r x + r^2 x = 73.5 x (rx)(r^2x)= 2744 You solve the two equations for x and r and you find two real solutions x=3.5 and r =4 or x=56 and r =1/4 Both yield the same 4 numbers.
how is it r^2?
let us suppose the three terms are \[{{a}\over{r}}, {a}, {ar}\] then \[{{a}\over{r}}+{a}+{ar} = 73.5\] and \[{{a}\over{r}}\times{a}\times{ar}=2744\] \[{a^3}=2744\] \[{a}=14\] now plug in this value in first equation \[{a}({{1}\over{r}}+{1}+{r}) = 73.5\] \[{14}({{1+r+r^2}\over{r}}) = 73.5\] \[{14}({{1+r+r^2}}) = 73.5r\] \[{14 +14r + 14r^2} = 73.5r\] \[{14r^2 -59.5r + 14} = 0\] now solve for r
.0O0. I bow down to your awesomeness! Thank you!!!!!
still having any difficulty ?
well, why do you assume that they are a/r, a, and ar?
because they are three consecutive numbers of the GP each having a common ratio in this case Common ration is r
and i was having multiplication so r gets cancelled and easy for calculation
GP?
Geometric Sequence or, Geometric Progression or, Geometric Series
oh! so like that other person was saying about the a+ar+ar^2 and stuff?
yeah both mean the same
SO I don't get how you are supposed to do this... I started factoring out things, but it's confusing...idk how to do...
Let x be the first element and r be the common ratio x + r x + r^2 x = 73.5 x (rx)(r^2x)= 2744 You solve the two equations for x and r and you find two real solutions x=3.5 and r =4 or x=56 and r =1/4 Both yield the same 4 numbers. Notice that the second equation yields r x =14 so x =14/r 14/r + 14 + 14 r = 73.5 Which yields r=4 or r = 1/4
uh...
okay.....
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