SUM OF THE FOUR CONSECUTIVE NUMBERS IN AN AP IS 32 AND THE RATIO OF THE PRODUCT OF THE FIRST AMD LAST TERMS TO THE PRODUCT OF TWO MIDDLE TERMS IS 7 RATIO 15 FIND THE NUMBER
let the numbers be a, a+d, a+2d and a+3d
a + a+d + a+2d + a+3d = 32
4a + 6d =32 ----> a = (16-3d)/2
a(a+3d)/( (a+d)(a+2d) ) = 7/15
(a^2 +3ad)/(a^2 + 3ad + 2d^2) = 7/15
15a^2 + 45ad = 7a^2 + 21ad + 14d^2
8a^2 + 24ad - 14d^2 = 0
8(16-3d)^2 /4 + 24d(16-3d)/2 - 14d^2 = 0
2(256 - 96d + 9d^2) + 12d(16 - 3d) - 14d^2 = 0
512 - 192d + 18d^2 + 192d - 36d^2 - 14d^2 = 0
-32d^2 = -512
finish it up, you will get 2 different sequences, they both work
To find the numbers, let's start by assuming that the first term of the arithmetic progression (AP) is 'a', and the common difference is 'd'.
The sum of four consecutive terms in an AP can be expressed as the sum of the first term, the second term, the third term, and the fourth term. So, we can write the equation as follows:
a + (a + d) + (a + 2d) + (a + 3d) = 32
Simplifying this equation will give us the value of 'a' in terms of 'd'. Let's proceed:
4a + 6d = 32 (combining like terms)
Dividing both sides of the equation by 2 will give us:
2a + 3d = 16 (equation 1)
Now, let's consider the ratio of the product of the first and last terms to the product of the two middle terms. The first term is 'a', and the last term is 'a + 3d'. The two middle terms are (a + d) and (a + 2d).
So, the ratio can be expressed as:
(a) * (a + 3d) / (a + d) * (a + 2d) = 7/15
Simplifying this equation will help us solve for 'a' and 'd'. Let's proceed:
(a^2 + 3ad) / (a^2 + 3ad + 2ad + 2d^2) = 7/15
Cross-multiplying will give us:
15(a^2 + 3ad) = 7(a^2 + 5ad + 2d^2)
Expanding both sides of the equation:
15a^2 + 45ad = 7a^2 + 35ad + 14d^2
Rearranging the terms:
15a^2 - 7a^2 + 45ad - 35ad = 14d^2
8a^2 + 10ad = 14d^2
Dividing both sides of the equation by 2 will give us:
4a^2 + 5ad = 7d^2 (equation 2)
Now, we have two equations:
2a + 3d = 16 (equation 1)
4a^2 + 5ad = 7d^2 (equation 2)
By solving these two equations simultaneously, we can find the values of 'a' and 'd'. Once we have the values of 'a' and 'd', we can find the four consecutive numbers in the AP.
Please note that this system of equations might have multiple solutions.