Question

Four numbers are in G.P. The sum of the first two numbers is 4, and the sum of the last two numbers is 36. Find the numbers.

Answer 1

`a/r^2, a/r, a, ar`

`a/r^2 + a/r` = 4

a + ar = 36

a(1 + r) = 36

a = `36/(1 + r)`

`a/r((1+ 1)/r)` = 4

`36/((1 + r)r) ((1 + r)/r)` = 4

`36/r^2` = 4

4r² = 36

r² = 9

r = `sqrt9`

r = ±3

If r = 3, a = `36/(1 + 3)`

a = 9

If r = −3, a = `36/(1 - 3)`

a = `(-36)/2`

a = −18

`a/r^2, a/r, a, ar`

If r = 3, a = 9

`9/3^2, 9/3, 9, 9 xx 3`

`9/9, 9/3, 9, 9 xx 3`

⇒ 1, 3, 9, 27

If r = −3, a = −18

`(-18)/-3^2, (-18)/(-3), -18, -18(-3)`

`(-18)/9, (-18)/(-3), -18, -18(-3)`

⇒ −2, 6, −18, 54