Question

Three numbers form a geometric progression. If the second term is increased by 2, the progression will become arithmetic and if, after this, the last term is increased by 9, then the progression will again become geometric. Find these three numbers
I found the first answer (4,8,16) what is the second answer? HELP

Answer 1

Answer:

0.16 -0.64 2.56

Step-by-step explanation:

the key is in bringing this to 2 equations with 2 variables. I focused on a1 (the first term of the sequence) and r (the common ratio of the original sequence).

a1×r = a2

a2×r = a3 = a1×r²

now, when we increase the second term by 2 we suddenly have an arithmetic sequence. that means that the differences between the terms must be the same.

a1×r + 2 - a1 = a3 - (a1×r + 2) = a1×r² - (a1×r + 2)

a1×r² - 2×a1×r + a1 - 4 = 0

and now, when we increase the third term by 9, we get a geometric sequence again. that means that the ratio of the terms must be the same.

(a1×r² + 9)/(a1×r + 2) = (a1×r + 2)/a1

(a1²×r² + 9×a1)/(a1×r + 2) = a1×r + 2

a1²×r² + 9×a1 = (a1×r + 2)² = a1²×r² + 4×a1×r + 4

4×a1×r - 9×a1 + 4 = 0

9×a - 4 = 4×a1×r

r = (9×a1 - 4)/(4×a1)

now we use that identity in the first equation :

a1×(9×a1 - 4)²/(4×a1)² - 2×a1×(9×a1 - 4)/(4×a1) + a1 - 4 = 0

(81×a1² - 72×a1 + 16)/(16×a1) - (9×a1 - 4)/2 + a1 - 4 = 0

81×a1² - 72×a1 + 16 - 8×a1×(9×a1 - 4) + 16×a1² - 64×a1 = 0

81×a1² - 72×a1 + 16 - 72×a1² + 32×a1 + 16×a1² - 64×a1 = 0

25×a1² - 104×a1 + 16 = 0

the general solution for a quadratic equation is

(-b ± sqrt(b² - 4ac))/(2a)

in our c case

a = 25

b = -104

c = 16

so,

a1 = (104 ± sqrt(104² - 4×25×16))/50

a1 = (104 ± sqrt(10816 - 1600))/50

a1 = (104 ± sqrt(9216))/50

a1 = (104 ± 96)/50

first a1 = (104 + 96)/50 = 200/50 = 4

and the corresponding r = (9×4 - 4)/(4×4) = 2

that was your original solution.

the second a1 = (104 - 96)/50 = 8/50 = 4/25 = 0.16

the corresponding r = (9×0.16 - 4)/(4×0.16) = -4

that gives us a the original sequence

0.16 -0.64 2.56

then adding 2 to the second term gives us

0.16 1.36 2.56 with the arithmetic difference of 1.2.

and then after adding 9 to the third term gives us the geometric sequence

0.16 1.36 11.56 with the common ratio of 8.5.

Answer 2

In this exercise we have to use the knowledge of geometric progression to find three specific numbers, in this way we can say that these numbers correspond to;

$0.16\\ -0.64\\ 2.56$

Then using the formula of the geometric progression we find that:

$a_1*r = a_2\\a_2*r = a_3 = a_1*r^2$

now, the differences between the terms must be the same:

$a1*r + 2 - a_1 = a_3 - (a_1*r + 2) = a_1*r^2 - (a_1*r + 2)\\a_1*r^2 - 2*a_1*r + a_1 - 4 = 0$

and now, when we increase the third term by 9,so we have:

$(a_1*r^2 + 9)/(a_1*r + 2) = (a_1*r + 2)/a_1\\(a_1^2*r^2 + 9*a_1)/(a_1*r + 2) = a_1*r + 2\\a_1^2*r^2 + 9*a_1 = (a_1*r + 2)^2 = a_1^2*r^2 + 4*a_1*r + 4\\4*a_1*r - 9*a_1 + 4 = 0\\9*a - 4 = 4*a_1*r\\r = (9*a_1 - 4)/(4*a_1)$

Now we use that identity in the first equation :

$a_1*(9*a_1 - 4)^2/(4×a_1)^2 - 2*a_1*(9*a_1 - 4)/(4*a_1) + a_1 - 4 = 0\\(81*a_1^2 - 72*a_1 + 16)/(16*a_1) - (9*a_1 - 4)/2 + a_1 - 4 = 0\\81*a_1^2 - 72*a_1 + 16 - 8*a_1*(9*a_1 - 4) + 16*a_1^2 - 64*a_1 = 0\\25a_1^2 - 104a_1 + 16 = 0$

The general solution for a quadratic equation is

$(-b + \sqrt{(b^2 - 4ac))/(2a)$

We have that:

• a = 25

• b = -104

• c = 16

so, put the numbers in the formula we find:

$a_1=4 \\a_1=0.16$

See more about geometric progress at brainly.com/question/14320920