let the four number be
a, a+d, a+2d , a+3d
then given:
a+5, a+d+6, a+2d+9, a+3d+15 are in GP
then (a+d+6)^2 = (a+5)(a+2d+9)
a^2 + d^2 + 36 + 2ad + 12a + 12d = a^2 + 2ad + 9a + 5a + 10d + 45
d^2 - 2a + 2d = 9
a = (d^2 + 2d - 9)/2
(a+2d+9)^2 = (a+d+6)(a+3d+15)
I get:
- 3a + d^2 + 3d - 9 = 0
sub a = (d^2 + 2d - 9)/2 into that and solve for d
I get 2 pairs of solution for a and d, one will not work, the other causes a 0 to show up in a GP, which is not possible.
I will leave it up to you to finish this.
four positive number are in a.p. If 5,6,9 and 15 are added respectively to these no.,we get a g.p. ,then common ratio of g.p. And commo difference of a.p. Is
3 answers
meant to say at the end:
I get 2 pairs of solution for a and d, one will work, the other causes a 0 to show up in a GP, which is not possible.
I get 2 pairs of solution for a and d, one will work, the other causes a 0 to show up in a GP, which is not possible.
Brainly is better