Since the deposits form a GP, we can say
a + ar + ar^2 = 65
Second condition:
to form an AP we would have the terms as
3a , 5ar and 3 ar^2
If they form an AP, then
5ar - 3a = 3ar^2 - 5ar
10ar - 3a = 3ar^2
divide each term by a
10r - 3 = 3r^2
3r^2 - 10r + 3 = 0
(3r - 1)(r - 3) = 0
r = 1/3 or r = 3
I will do the case of r = 1/3, you do the case when r = 3
They both work
if r = 1/3, in the first equation:
a + a/3 + a/9 = 65
times 9
9a + 3a + a = 585
13a = 585
a = 45
So the first two deposits are a and ar, which would be
45 and 15
For three consecutive months, a person deposited some amount of money on the first day of ech montg in a small saving fund. These three successive amounts in the deposiy, the total value of which is ksh. 65,form a GP if the 2 extreme amounts be multiplied each by 3 abd the mean by 5, the product forms an AP. Find the amounts in the first and secont deposit
2 answers
GP: a+ar+ar^2=65
AP: 3a, 5ar, 3ar^2
According to AP,
5ar-3a=3ar^2-5ar
10ar-3a=3ar^2
a(10r-3) =3ar^2
10r-3=3r^2
3r^2-10r+3=0
(3r-1) (r- 3) =0
r=1/3 or r=3
Substitute in GP, (taking r=3)
a+ar+ar^2=65
a+3a+9a=65
13a=65
a=5
So the first two deposits a and ar, which would be 5 and 15
AP: 3a, 5ar, 3ar^2
According to AP,
5ar-3a=3ar^2-5ar
10ar-3a=3ar^2
a(10r-3) =3ar^2
10r-3=3r^2
3r^2-10r+3=0
(3r-1) (r- 3) =0
r=1/3 or r=3
Substitute in GP, (taking r=3)
a+ar+ar^2=65
a+3a+9a=65
13a=65
a=5
So the first two deposits a and ar, which would be 5 and 15
0 answers