The sum of three numbers in G.P. is 38 and their product is 1728; find them.
Answers
Answered by
2
Take the numbers as a ar ar^2
The product is (ar)^3=(12)^3
Therefore ar=12
Therefore a=12/r
Now sum is 38
a+ar+ar^2=38
Now substitute a=12/r
You get(12+12r+12r^2)/r=38
So 12+12r+12r^2=38
So 12r^2-26r+12=0
i.e. 6r^2-13r+6=0
There fore r=3/2 or 3
We reject 3
We get G.P as 18,12,8
The product is (ar)^3=(12)^3
Therefore ar=12
Therefore a=12/r
Now sum is 38
a+ar+ar^2=38
Now substitute a=12/r
You get(12+12r+12r^2)/r=38
So 12+12r+12r^2=38
So 12r^2-26r+12=0
i.e. 6r^2-13r+6=0
There fore r=3/2 or 3
We reject 3
We get G.P as 18,12,8
Answered by
0
Answer:
Let a/r, a, and ar be the three numbers in GP.
Sum, a/r + a + ar = 38 …(i)
Product, (a/r)a(ar) = 1728
a³= 1728
Taking cube root
a = 12
Substitute a in (i)
(12/r) + 12 + 12r = 38
(12/r) + 12r = 26
((1/r) + r) = 26/12
(r² + 1)/ r = 13/6
6r²-13r+6 = 0
Solving using the quadratic formula, we get
r = 2/3or 3/2
The numbers will be 18, 12, 8 or 18, 12, 8.
The greatest number is 18.
Similar questions