Question

In a geometric progression the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2. Find the three terms.​

Answer 1

this is your answer

Answer 2

The three terms are either 2, 3, 9/2 or 9/2, 3, 2.

Step-by-step explanation:

Let three consecutive terms of a GP are a/r, a, ar.

Product of three consecutive terms is 27.

$\dfrac{a}{r}\times r\times ar=27$

$a^3=27$

$a=3$

The sum of the product of two terms taken at a time is 57/2.

$\dfrac{a}{r}\cdot a+a\cdot ar+\dfrac{a}{r}\cdot ar=\dfrac{57}{2}$

$\dfrac{a^2}{r}+a^2r+a^2=\dfrac{57}{2}$

Substitute a=3,

$\dfrac{3^2}{r}+3^2r+3^2=\dfrac{57}{2}$

$\dfrac{9}{r}+9r+9=\dfrac{57}{2}$

$\dfrac{9+9r^2}{r}=\dfrac{57}{2}-9$

$\dfrac{9+9r^2}{r}=\dfrac{39}{2}$

$18+18r^2=39r$

$18r^2-39r+18=0$

$(2r-3)(9r-6)=0$

$r=\dfrac{3}{2},\dfrac{2}{3}$

For a=3 and r=3/2 the three terms are 2, 3, 9/2.

For a=3 and r=2/3 the three terms are 9/2, 3, 2.

Therefore, the three terms are either 2, 3, 9/2 or 9/2, 3, 2.

#Learn more

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