Question

The sum of three numbers in G.P. is 38, and their product is 1728. Find each number?

Answer 1

Hint: For solving this problem, first we must let three numbers in a GP be . Now, multiplying all the numbers we obtain the product in a single variable. Hence, we evaluate all the unknown variables to obtain the solution.

Complete step-by-step answer:
When some numbers have a definite recurring pattern then they form a sequence or series. If we obtain a series for a given set of numbers then various operations can be easily performed using different formulas.
According to the question we are given that the sum of three numbers in GP is 38 and the product of these three numbers is 1728. Let the three numbers in GP be $\dfrac{a}{r},a,ar$ where a is a constant number and r is the ratio of geometric progression.
Now, using the problem statement we get
$\begin{aligned} & \dfrac{a}{r}+a+ar=38 \\\ & a\left( \dfrac{1}{r}+1+r \right)=38\ldots (1) \\\ & \dfrac{a}{r}\times a\times ar=1728 \\\ & {{a}^{3}}=1728={{12}^{3}} \\\ & \therefore a=12\ldots (2) \\\ \end{aligned}$
Now, putting the value of a from equation (2) to equation (1) we get,
$\begin{aligned} & 12\left( {{r}^{2}}+r+1 \right)=38r \\\ & 6\left( {{r}^{2}}+r+1 \right)=19r \\\ & 6{{r}^{2}}-13r+6=0 \\\ \end{aligned}$
Now, by the middle term splitting method, factorising the equation to get r.
$\begin{aligned} & 6{{r}^{2}}-9r-4r+6=0 \\\ & 3r(2r-3)-2(2r-3)=0 \\\ & (3r-2)(2r-3)=0 \\\ & \therefore r=\dfrac{2}{3},\dfrac{3}{2} \\\ \end{aligned}$
For a = 12 and $r=\dfrac{2}{3}$, the corresponding numbers are 18, 12, 8.
For a = 12 and $r=\dfrac{3}{2}$, the corresponding numbers are 8, 12, 18.
Note: The key step for solving this problem is the knowledge of the number system and particularly sequence and series. The basic idea geometric progression and the assumption for three terms in G.P. is good enough to solve both the parts. This knowledge is helpful in solving complex problems.