Question

The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP.
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Answer 1

Answer:

• The required increasing G.P will be 12 , 18 & 27 respectively .

Step-by-step explanation:

According to the Question

It is given that,

• Product of three increasing numbers in GP is 5832

• If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP.

Let the first three terms of the given G.P be $\sf \frac{a}{r} , a \; and \; ar$ respectively .

Product of given G.P = 5832

↠ a/r × a × ar = 5832

↠ a × a × a = 5832

↠ a³ = 5832

↠ a³ = 18 × 18 × 18

↠ a³ = 18³

↠ a = 18 .

And, also it is given that a/r , (a+6) & (ar+9) are in A.P (Given)

$\dashrightarrow\sf\; 2\; (a+6) = \frac{a}{r} + ar + 9 \\\\\\\dashrightarrow\sf\; 2\; (18+6) = \frac{18}{r} + 18r + 9 \\\\\\\dashrightarrow\sf\; 2\; (24) = \frac{18}{r} + 18r + 9 \\\\\\\dashrightarrow\sf\; 48 -9 = \frac{18}{r} + 18r \\\\\\\dashrightarrow\sf\; 39 = \frac{18 + 18r^{2}}{r} \\\\\\\dashrightarrow\sf\; 39r = 18 + 18r^{2} \\\\\\\dashrightarrow\sf\; 18r^{2} -39r + 18 = 0\\\\\\\dashrightarrow\sf\; 6r^2 -13r + 6 = 0\\\\\\\dashrightarrow\sf\; 6r^2 - 9r - 4r + 6 = 0$

$\dashrightarrow\sf\; 3r(2r-3) -2(2r-3) = 0\\\\\\\dashrightarrow\sf\; (2r-3) (3r-2) = 0\\\\\\\dashrightarrow\sf\; r = \frac{3}{2} \; r = \frac{2}{3}$

As it is given that the G.P are in increasing order .

So we will take here r = 3/2

So, the required increasing G.P will be

$\dashrightarrow\sf\; \frac{a}{r} = \frac{18}{\frac{3}{2} } = 18 \times\frac{2}{3} = 6\times\;2 = 12\\\\\\\dashrightarrow\sf\; a = 18 \\\\\\\dashrightarrow\sf\; ar = 18\times\; \frac{3}{2} = 9\times\; 3 =27 \\\\\\\boxed{\bf{{Hence,\; the\; required\; increasing \; GP \; are\; 12, 18 \; and \; 27}}}$

Answer 2

Your answer is on the attachment..