Three numbers are in G.P. whose sum is 70. if the extremes be each multiplied by 4 and the means by 5, they will be in AP. find the numbers .
To find the numbers, we can start by solving the problem step by step.
Let's assume the three numbers in the geometric progression (G.P.) are a, ar, and ar^2, where r is the common ratio.
According to the given information:
Sum of the three numbers = 70
a + ar + ar^2 = 70 --------------- (1)
We are also told that when the extremes are multiplied by 4 and the means by 5, they will form an arithmetic progression (A.P.). Therefore:
4a, ar^2, and 5ar are in arithmetic progression.
So, the common difference (d) of the arithmetic progression (A.P.) can be found by subtracting consecutive terms:
ar^2 - 4a = 5ar - ar^2
ar^2 - 5ar + 4a = 0 --------------- (2)
Now we have two equations (1) and (2) that we can solve simultaneously to find the values of a and r.
From equation (2), we can factorize it:
a(r^2 - 5r + 4) = 0
This gives us two cases:
1) a = 0:
If a = 0, then equation (1) becomes:
0 + 0 + ar^2 = 70
ar^2 = 70
Since a cannot be zero, we move to the second case.
2) (r^2 - 5r + 4) = 0:
Solving this quadratic equation, we can use factoring or the quadratic formula.
(r - 4)(r - 1) = 0
This gives two possible values for r:
r = 4 or r = 1
Case 1: r = 4
Using r = 4 in equation (1), we get:
a + 4a + 16a = 70
21a = 70
a = 70/21 = 10/3
Thus, when r = 4, the three numbers in the geometric progression (G.P.) are:
a = 10/3, ar = (10/3) * 4 = 40/3, and ar^2 = (10/3) * 4^2 = 160/3
Case 2: r = 1
Using r = 1 in equation (1), we get:
a + a + a = 70
3a = 70
a = 70/3
Thus, when r = 1, the three numbers in the geometric progression (G.P.) are:
a = 70/3, ar = (70/3) * 1 = 70/3, and ar^2 = (70/3) * 1^2 = 70/3
So, we have found two possible sets of numbers that satisfy the given conditions:
1) (10/3, 40/3, 160/3)
2) (70/3, 70/3, 70/3)
a+ar+ar^2 = 70
5ar-4a = 4ar^2-5ar
GP is 10,20,40
AP is 40,100,160