The sum of 3 numbers in GP is 38 and their product is 1728. The greatest number of them is ....
Answers
Answer:
18
Step-by-step explanation:
ANSWER
Let the required three numbers of G.P. are r′aaand ar. Then, their sum =ra+a+ar=38
⇒a(r1+r+r2)=38 ....... (i)
product =ra×a×ar=1728
⇒a3=(12)3
∴a=12 ........ (ii)
Substitute the value of a, in equation (i), we get
∴12×(r1+r+r2)=38
⇒6+6r+6r2=19r
⇒6r2−13r+6=0
⇒(3r−2)(2r−3)=0∴r=32 or 23
Hence, the required numbers are 18, 12, 8 or 8, 12, 18
∴ Greatest number = 18.
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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.