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Certainly! Let's approach this problem systematically. Since we are dealing with a geometric progression (GP), let's represent the three numbers as arar, ar2ar2, and ar3ar3, where aa is the first term and rr is the common ratio.
We're given that the product of these three numbers is 1728, so we can form the equation:
(ar)(ar2)(ar3)=1728(ar)(ar2)(ar3)=1728
a3r6=1728a3r6=1728
Similarly, we're also given that the sum of these three numbers is 38, so:
ar+ar2+ar3=38ar+ar2+ar3=38
Now, let's solve these equations step by step. Firstly, let's rewrite the product equation:
a3r6=1728a3r6=1728
a3r6=26×33a3r6=26×33
(ar)3×(ar2)3=26×33(ar)3×(ar2)3=26×33
Now, we can see that ar=2ar=2 and ar2=3ar2=3, because 2 and 3 are the prime factors of 1728 and they are consecutive powers. So, let's solve for aa and rr:
From ar=2ar=2, we get a=2ra=r2
From ar2=3ar2=3, we get a=3r2a=r23
Equating these expressions for aa:
2r=3r2r2=r23
2r=32r=3
r=32r=23
Now that we have found rr, we can find aa:
ar=2ar=2
a×32=2a×23=2
a=43a=34
So, the three numbers are 4334, 2, and 3.
Therefore, as an experienced tutor registered on UrbanPro, I have demonstrated how to find three numbers in GP whose product is 1728 and sum is 38. UrbanPro is indeed a great platform for seeking online coaching and tuition services, where experienced tutors can provide comprehensive assistance in various subjects.
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