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Exercise 5.2 · Q4

Q.The product of three increasing numbers in GP is 58325832. If we add 66 to the second number and 99 to the third number, then the resulting numbers form an AP. Find the numbers in GP.

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Let the three GP numbers be ar,a,ar\dfrac ar,a,ar; their product pins down aa directly, and the AP condition on the modified numbers gives a quadratic for rr.

Step 1. Set up the GP and use the product. Let the numbers be ar, a, ar\dfrac ar,\ a,\ ar. Product =a3=5832=183⇒a=18=a^3=5832=18^3\Rightarrow a=18.

Step 2. Set up the AP condition. Adding 66 to the second and 99 to the third: ar, a+6, ar+9\dfrac ar,\ a+6,\ ar+9 form an AP, so

2(a+6)=ar+ar+9.2(a+6) = \frac ar+ar+9.

Step 3. Substitute a=18a=18. 2(24)=18r+18r+9⇒48=18r+18r+9⇒39=18r+18r2(24)=\dfrac{18}r+18r+9 \Rightarrow 48=\dfrac{18}r+18r+9 \Rightarrow 39=\dfrac{18}r+18r.

Step 4. Clear the denominator. Multiply by rr: 39r=18+18r2⇒18r2−39r+18=0⇒6r2−13r+6=039r=18+18r^2 \Rightarrow 18r^2-39r+18=0 \Rightarrow 6r^2-13r+6=0 (dividing by 3).

Step 5. Solve the quadratic. r=13±169−14412=13±512⇒r=32r=\dfrac{13\pm\sqrt{169-144}}{12}=\dfrac{13\pm5}{12} \Rightarrow r=\dfrac32 or r=23r=\dfrac23. …

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