Question

the sum of 4 consecutive nos. in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms is 7:15.Find the nos.

Answer 1

yeh to aaj math ke exam me aya tha na....
mujhe nhi aata

Answer 2

Given: Sum of A.P = 32
Sum = n/2* (A1 + An)... n= no: of terms, A1= 1st term, An= nth term
Here n = 4
32 = 4/2*(A1+A4)
A1 + A4 = 16....(a)

Given: A1*A4/A2*A3 = 7/15
A2 = A1 + d
A3 = A1 + 2d
A4 = A1 + 3d
Substitute these values:
A1*(A1+3d)/(A1+d)*(A1+2d) = 7/15
A1^2 + 3dA1 / (A1^2 + 3dA1 + 2d^2) = 7/15
15*(A1^2 + 3dA1) = 7*(A1^2 + 3dA1 + 2d^2)
15A1^2 + 45dA1 = 7A1^2 + 21dA1 + 14d^2
8A1^2 + 24dA1 - 14d^2 = 0
8A1^2 - 4dA1 + 28dA1 - 14d^2 = 0
4A1*(2A1 - d) + 14d*(2A1 - d) = 0
[4A1 + 14d] * [2A1 - d] = 0
4A1 = -14d   OR 2A1 = d
2A1 = -7d    OR 2A1 = d
Since we need a positive A.P (coz sum of 1st and 4th = 32), we have to consider the case: 2A1 = d....(b)
(a)===A1 + A4 = 16 
A1 + A1 + 3d = 16
2A1 + 3d = 16
d + 3d = 16
d = 4
2A1 = d
A1 = d/2 = 2
Hence, A.P is:
2, 6, 10, 14
Sum = 32
Ratio of product of 1st and 4th to 2nd and 3rd:
2*14/6*10 = 28/60 = 7/15 = 7:15
All conditions verified.
Hence the nos. are 2,6,10,14
Hope it helps.