Question

Find X so that X+6,X+12 and X+15 are consecutive terms of a geometric progression.​

Answer 1

The value of X is -18

Step-by-step explanation:

Given that X+6, X+12 and X+15 are consecutive numbers of a geometric progression

To find the value of X :

Since the given numbers are in GP

Let $a_1=X+6$ , $a_2=X+12$ and $a_3=X+15$

The common ratio $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}$

• Now substitute the values we have
• $r=\frac{X+12}{X+6}=\frac{X+15}{X+12}$
• $\frac{X+12}{X+6}=\frac{X+15}{X+12}$
• $(X+12)\times (X+12)=(X+15)\times (X+6)$
• $(X+12)^2=X^2+6X+15X+90$ ( by using distributive property )

• $X^2+24X+144=X^2+6X+15X+90$ ( using the identity $(a+b)^2=a^2+2ab+b^2$ )
• $X^2+24X+144-X^2-6X-15X-90=0$  ( adding the like terms )
• $3X+54=0$
• $3X=-54$
• $X=-\frac{54}{3}$
• $X=-18$

Therefore the value of X is -18

Answer 2

Answer:

Step-by-step explanation:

The value of X is -18

Step-by-step explanation:

Given that X+6, X+12 and X+15 are consecutive numbers of a geometric progression

To find the value of X :

Since the given numbers are in GP

Let , and

The common ratio

Now substitute the values we have

( by using distributive property )

( using the identity )

 ( adding the like terms )

Therefore the value of X is -18