three numbers whose sum is 129 in GP were the extreme are multiplied by 12 and the mean is multiplied by 37 they result in AP find the numbers
Answers
Answer:
Let the numbers in GP be a, ar and ar^2.
Their sum = a+ar+ar^2 = 129, or
a(1+r+r^2) = 129 …(1)
4a, 5ar and 4ar^2 are in AP. Or
5ar-4a = 4ar^2–5ar, or
a(5r-4) = ar(4r-5), or
5r-4 = 4r^2-5r, or
4r^2-10r+4 = 0
2r^2-5r+2 = 0
(2r-1)(r-2)=0
r = 1/2 or 2
a+ar+ar^2 = 129 …(1)
With r = 1/2, (1) becomes a(1+0.5+0.25) = 129, or
a = 129/1.75 = 73.71428571
The three terms of the GP are: 73.71428571, 36.85714286, 18.42857143
With r =2, (1) becomes a(1+2+4) = 129, or
a = 129/7 = 18.42.
The three terms of the GP are: 18.42857143, 36.85714286, 73.71428571. or 129/7, 258/7 and 516/7. Answer.
Step-by-step explanation:
please mark me brain list please please please please please please please please please please please please please
Answer:
Let the numbers in GP be a, ar and ar^2.
Their sum = a+ar+ar^2 = 129, or
a(1+r+r^2) = 129 …(1)
4a, 5ar and 4ar^2 are in AP. Or
5ar-4a = 4ar^2–5ar, or
a(5r-4) = ar(4r-5), or
5r-4 = 4r^2-5r, or
4r^2-10r+4 = 0
2r^2-5r+2 = 0
(2r-1)(r-2)=0
r = 1/2 or 2
a+ar+ar^2 = 129 …(1)
With r = 1/2, (1) becomes a(1+0.5+0.25) = 129, or
a = 129/1.75 = 73.71428571
The three terms of the GP are: 73.71428571, 36.85714286, 18.42857143
With r =2, (1) becomes a(1+2+4) = 129, or
a = 129/7 = 18.42.
The three terms of the GP are: 18.42857143, 36.85714286, 73.71428571. or 129/7, 258/7 and 516/7
this is the answer of your question hope it will help you