Question

Find four numbers that form a geometric progression such that the third term is greater than the first by 12 and the fourth is greater than the second by 36.

Answer 1

ar^2 = a+12
ar^3 = ar+36

Answer 2

The n-th term of a geometric sequence with initial value a and common ratio r is given by:

an = a * r ^ ( n - 1 )

a is the first term
r is common ratio

In this case:

a1 = a * r ^ ( 1 - 1 ) = a * r ^ 0 = a * 1 = a

a2 = a * r ^ ( 2 - 1 ) = a * r ^ 1 = a * r

a3 = a * r ^ ( 3 - 1 ) = a * r ^ 2

a4 = a * r ^ ( 4 - 1 ) = a * r ^ 3


The third term is greater than the first by 12 mean:

a3 = a1 + 12

The fourth is greater than the second by 36 mean:

a4 = a2 + 36

So:

a3 = a1 + 12

a * r ^ 2 = a + 12


a4 = a2 + 36

a * r ^ 3 = a * r + 36


Now you must solve system of 2 equations with 2 unknow:

a * r ^ 2 = a + 12

a * r ^ 3 = a * r + 36


a * r ^ 2 = a + 12 Subtract a to both sides

a * r ^ 2 - a = a + 12 - a

a * r ^ 2 - a = 12

a * ( r ^ 2 - 1 ) = 12


a * r ^ 3 = a * r + 36 Subtract a * r to both sides

a * r ^ 3 - a * r = a * r + 36 - a * r

a * r ^ 3 - a * r = 36

a * r * ( r ^ 2 - 1 ) = 36

Your system become:

a * ( r ^ 2 - 1 ) = 12

a * r * ( r ^ 2 - 1 ) = 36


a * ( r ^ 2 - 1 ) = 12 Divide both sides by a

r ^ 2 - 1 = 12 / a


a * r * ( r ^ 2 - 1 ) = 36 Divide both sides by a * r

r ^ 2 - 1 = 36 / a * r


r ^ 2 - 1 = r ^ 2 - 1

12 / a = 36 / a * r Multiply both sides by a

12 * a / a = 36 * a / a * r

12 = 36 / r Multiply both sides by r

12 * r = 36 * r / r

12 r = 36 Divide both sides by 12

r = 36 / 12

r = 3


Replace this value in equation:

a * ( r ^ 2 - 1 ) = 12


a * ( 3 ^ 2 - 1 ) = 12

a * ( 9 - 1 ) = 12

a * 8 = 12

8 a = 12 Divide both sides by 8

a = 12 / 8

a = 4 * 3 / ( 4 * 2 )

a = 3 / 2


a = 3 / 2 , r = 3 so:

a1 = a

a1 = 3 / 2


a2 = a * r

a2 = ( 3 / 2 ) * 3 = 9 / 2


a3 = a * r ^ 2

a3 = ( 3 / 2 ) * 3 ^ 2 = ( 3 / 2 ) * 9 = 27 / 2


a4 = a * r ^ 3

a4 = ( 3 / 2 ) * 3 ^ 3 = ( 3 / 2 ) * 27 = 81 / 2


Your geometric progression:

3 / 2 , 9 / 2 , 27 / 2 , 81 / 2

Proof:

a3 = a1 + 12

27 / 2 = 3 / 2 + 12

27 / 2 = 3 / 2 + 12 * 2 / 2

27 / 2 = 3 / 2 + 24 / 2


a4 = a2 + 36

81 / 2 = 9 / 2 + 36

81 / 2 = 9 / 2 + 36 * 2 / 2

81 / 2 = 9 / 2 + 72 / 2

Answer 3

Nice work. I'd have changed things a bit here:

Your system become:

a * ( r ^ 2 - 1 ) = 12
a * r * ( r ^ 2 - 1 ) = 36

Divide the bottom by the top to get

r = 36/12 = 3
a*8 = 12
a = 3/2