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
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
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
Well, let's start solving this puzzle! Since the three successive amounts form a geometric progression, let's assume the first deposit is "a," the common ratio is "r," and the second deposit is "ar."
So, the three deposits can be written as:
First deposit = a
Second deposit = ar
Third deposit = ar^2
We know that the sum of these three deposits is equal to 65:
a + ar + ar^2 = 65
Now, according to the second part of the information, if we multiply the two extreme amounts by 3 (a and ar) and the mean by 5 (ar), their product forms an arithmetic progression. Mathematically, it can be expressed as:
3a * 5ar = 5 * (3ar)^2
Simplifying this equation, we get:
15a^2r^2 = 45a^2r^2
Canceling out the common terms, we're left with:
15 = 45
Oops! It seems I made a mistake somewhere. Let's have another go at it!
Why did the math book go to the therapist? Because it had too many problems!
Okay, let's get back to solving the question.
Do you have any other information or constraints that can help us narrow down the solution?
Let's consider the three consecutive deposits as a geometric progression (GP):
Let the first deposit be a.
The second deposit can be calculated as a * r, where r is the common ratio of the GP.
The third deposit can be calculated as a * r^2.
We are given that the sum of the three deposits is 65 Ksh:
a + a * r + a * r^2 = 65 ------ (Equation 1)
We are also given that if we multiply the two extreme amounts (a and a * r^2) by 3 each, and the mean (a * r) by 5, the product forms an arithmetic progression (AP).
3 * a * 3 * a * r^2 = 5 * a * r ------ (Equation 2)
Simplifying Equation 2, we get:
9 * a^2 * r^2 = 5 * a * r
Dividing both sides of the equation by a * r (assuming a and r are not zero), we get:
9 * a * r = 5
Simplifying further, we find:
a * r = 5/9 ------ (Equation 3)
Now, we can solve Equations 1 and 3 simultaneously to find the values of a and r. Since it's a quadratic equation, we'll first simplify it.
Using Equation 3, we can substitute the value of r in terms of a into Equation 1:
a + a * (5/9) + a * (5/9)^2 = 65
Simplifying this equation, we get:
9a + 5a + (25/9)a = 585
Multiplying by 9 to get rid of the denominator, we have:
81a + 45a + 25a = 5265
Simplifying further, we find:
151a = 5265
Dividing both sides by 151:
a = 5265/151
a ≈ 34.78
Now, substituting the value of a back into Equation 3:
34.78 * r = 5/9
Simplifying:
r = (5/9) / 34.78
r ≈ 0.04192
Therefore, the first deposit (a) is approximately 34.78 Ksh and the second deposit (a * r) is approximately 1.46 Ksh.
To solve this problem, we need to set up equations based on the information given.
Let's assume the first deposit amount is x. Since the three successive amounts form a geometric progression, the second deposit amount would be x * r, and the third deposit amount would be x * r^2, where r is the common ratio.
According to the information provided, the sum of these three amounts is Ksh. 65. Therefore, we can write the equation:
x + x * r + x * r^2 = 65
Now, we are told that if we multiply the two extreme amounts (the first and third deposits) by 3 and the mean amount (the second deposit) by 5, the resulting values form an arithmetic progression. Mathematically, this can be expressed as:
3x, 5(x * r), and 3x * r^2
To check if these values form an arithmetic progression, we can set up another equation:
5(x * r) - 3x = 3x * r^2 - 5(x * r)
Simplifying this equation, we get:
5xr - 3x = 3x * r^2 - 5xr
Combining like terms:
10xr - 6x = 3x * r^2
Dividing both sides by x:
10r - 6 = 3r^2
Rearranging this equation, we get:
3r^2 - 10r + 6 = 0
Now we have two equations:
1) x + x * r + x * r^2 = 65
2) 3r^2 - 10r + 6 = 0
We can solve these equations simultaneously to find the values of x and r, which will give us the first and second deposit amounts.
Let's solve equation 2) first by factoring or using the quadratic formula. After solving for r, we can substitute the value of r into equation 1) to find x.
Once we have the values of x and r, we can find the first and second deposit amounts by multiplying x by r and x by r^2, respectively.