The product of three consecutive term is x,y,z in an GP is 5832 if the sum of three term - Maths -

Question

The product of three consecutive term is x,y,z in an GP is 5832
if the sum of three term is 57 then what is the value of 3x+2y+3z
?

Ankush Jain · Accepted Answer

Given the product of three consecutive terms x, y and z in g.p. = 5832

We know that if there are three consecutive terms in g.p then they are taken as:

$\frac{a}{r}, a a n d a r$

According to the given condition,

$\frac{a}{r} \times a \times a r = 5832 \Rightarrow a^{3} = 5832 \Rightarrow a^{3} = 18^{3} \Rightarrow a = 18$

Also, it is given that sum of three terms = 57

$\Rightarrow \frac{a}{r} + a + a r = 57 \Rightarrow \frac{18}{r} + 18 + 18 r = 57 \Rightarrow \frac{18 + 18 r + 18 r^{2}}{r} = 57 \Rightarrow 18 + 18 r + 18 r^{2} = 57 r \Rightarrow 18 r^{2} - 39 r + 18 = 0 \Rightarrow 18 r^{2} - 27 r - 12 r + 18 = 0 \Rightarrow 9 r (2 r - 3) - 6 (2 r - 3) = 0 \Rightarrow (9 r - 6) (2 r - 3) = 0 \Rightarrow 3 (3 r - 2) (2 r - 3) = 0 \Rightarrow r = \frac{2}{3}, \frac{3}{2}$

When $r = \frac{2}{3}$ , then

$x = \frac{a}{r} = \frac{18}{\frac{2}{3}} = 27,$

y = a = 18 and z = ar = $18 \times \frac{2}{3} = 12$

When $r = \frac{3}{2}$ , then

$x = \frac{a}{r} = \frac{18}{\frac{3}{2}} = 12,$

y = a = 18 and z = ar = $18 \times \frac{3}{2} = 27$

Therefore, for $r = \frac{2}{3}$ ,

3x + 2y + 3z = 3(27) + 2(18) + 3(12) = 153.

Similarly, for $r = \frac{3}{2}$ ,

3x + 2y + 3z = 3(12) + 2(18) + 3(27) = 153.

Hence, the values of 3x + 2y + 3z = 153 for both $r = \frac{2}{3}, \frac{3}{2}$

The product of three consecutive term is x,y,z in an GP is 5832. if the sum of three term is 57 then what is the value of 3x+2y+3z ?