Arithmetic sequence to Geometric sequence?

Question

Three numbers form an arithmetic sequence having a common difference of 4. If the first number is increased by 2, the second number by 3, and the 3rd number by 5, the resulting numbers form a geometric sequence. Find the original numbers.


Please help me. I really tried solving it using different equations but still, i cant get an answer. Please?

10 pts for the best answer. Thanks

Answer 1

let the middle number of the arithmetic sequence be x. then the 3 of them are x-4, x, and x+4. after the increases, they are x-2, x+3, and x+9. if they form a geometric sequence, we'll have equal ratios:

(x-2)/(x+3) = (x+3)/(x+9)
(x-2)(x+9) = (x+3)²
x² + 7x - 18 = x² + 6x + 9
7x - 18 = 6x + 9
x = 27.

check it:
23, 27, 31 are the arithmetic sequence.
25, 30, 36 are geometric since
25/30 = 5/6
30/36 = 5/6

Answer 2

Let the first number be x
the second and third numbers will be x+4 and x+8

According to second condition the new numbers come out to be
x+2, x+4+3 and x+8+5 or x+2, x+7 and x+13.
These new numbers are in G.P.
It means
(x+2)/(x+7)=(x+7)/(x+13)

To solve for x, cross multiply
(x+2)*(x+13)=(x+7)*(x+7)
or x^2+15x+26=x^2+14x+49
or15x-14x=49-26
or x=23

Therefore three numbers are 23, 27 and 31

Answer 3

x= first number of sequence
x+4 = 2nd
x+8 = 3rd
for the second sequence
x+2 = first
x+7 = second
x+13 = third

for a geometric sequence there has to be a common multiplier
a(x+2)=x+7
a(x+7)=x+13
or
ax + 2a = x+7
ax +7a = x+13
subract the two
2a-7a=7-13
5a=6
a=1.2
and since 1.2x+2(1.2) = x+7
.2x=4.6
x=23
then the original #'s are 23,27,31

Answer 4

Ah, nevermind. Time for me to wander off for an other cup of coffee, I think.