Question

The sum of 3 numbers in a G.P. is 19, and the sum of their squares is 133. Find the numbers.

Answer 1

`a/r`, a, ar

`a/r + a + ar` = 19

`a(1/r + 1 + r)`  = 19

a(1 + r + r²)  = 19r

a = `(19r)/((1 + r + r^2))`   ....(1)

`(a/r)^2 +  a^2 + (ar)^2` = 133

⇒ `a^2/r^2 +  a^2 + a^2r^2` = 133

`a^2(1/r^2 + 1 + r^2)` = 133

`a^2(1 + r^2 + r^4) = 133r^2`

⇒ `a^2 = (133r^2)/((r^2 + 1 - r)(r^2 + r + 1))`   ....(2)

`((19)/(1 + r + r^2))^2 = (133)/((r^2 + 1 - r)(r^2 + r + 1))`

361(1 + r² − r) = 133(1 + r + r²)

`r = 3/2, r = 2/3`

`r = 3/2`

a = `19/(1 + r + r^2)`

= `19/(2/3 + 4/9 + 1)`

= `(19 xx 9)/19`

= 9

9, 6, 4

`r = 2/3`

= `19/(3/2 + 9/4 + 1)`

= `(19 xx 8)/19`

= 8

4, 6, 9