Question

Three numbers, of which the third is equal to 36, form a geometric progression. If 36 is replaced with 20, then the three numbers form an arithmetic progression. Find these three numbers.

Answer 1

Answer: The answer is either 4, 12, 36 or  100, 60, 36.

Step-by-step explanation: Let the first two numbers be 'a' and 'b'.

According to the question, a, b and 36 forms a geometric series. So, we have

$\dfrac{b}{a}=\dfrac{36}{b}\\\\\Rightarrow b^2=36a.$

Also, if 36 is replaced by 20, then a, b, 20 will form an arithmetic series. So,

$b-a=20-b\\\\\Rightarrow 2b=20+a\\\\\Rightarrow 2b=20+\dfrac{b^2}{36}\\\\\Rightarrow 72b=720+b^2\\\\\Rightarrow b^2-72b+720\\\\\Rightarrow b^2-60b-12b+720=0\\\\\Rightarrow b(b-60)-12(b-60)=0\\\\\Rightarrow (b-12)(b-60)\\\\\Rightarrow b=12,~~b=60.$

Therefore,

$a=\dfrac{12^2}{36}=4,~~a=\dfrac{60^2}{36}=100.$

Thus, the three numbers are either 4, 12, 36 or  100, 60, 36.

Answer 2

Answer:

100, 60, 36

or

4, 12, 36

Step-by-step explanation:

Three numbers, of which the third is equal to 36, form a geometric progression

let em' be a, ar n ar^2=36

If 36 is replaced with 20, then the three numbers form an arithmetic progression

difference in 1st 2 nos is ar-a

so third no. is a+2(ar-a)=a+2ar-2a=2ar-a=20

2 eqns w 2 unknowns:

ar^2=36 or a=36/r^2

2ar-a=20 or a(2r-1)=20

substitute 1st eqn into 2nd eqn

(36/r^2)(2r-1)=20

36(2r-1)=20r^2

9(2r-1)=5r^2

5r^2-18r+9=0

(5r-3)(r-3)=0

r=3/5 or 3

a=36/r^2

=100 or 4

numbers are 100, 60, 36

or

4, 12, 36