Question

find five consecutive terms in A.P whose sum is 120 and ratio of the products of the first and the last term to the product of the second and the fourth term is 20 : 21

Answer 1

let the terms of the A.P. be


a1

a2 = a1 + d

a3 = a1 + 2d

a4 = a1 + 3d

a5 = a1 + 4d


the sum of the 5 consecutive terms is


S5 = (5/2)(a1 + a5)


120 = (5/2)(a1 + a1 + 4d)


5(a1 + 2d) = 120


a1 + 2d = 24


a1 = 24 - 2d


the ratio of the product of the first n d last terms to the product of 2nd n 4th term is 20:21


(a1*a5) : (a2*a4) = 20 : 21


21a1(a1 + 4d) = 20(a1 + d)(a1 + 3d)


plug a1 = 24 - 2d into the last equaton


21(24 - 2d )(24 - 2d + 4d) = 20(24 - 2d + d)(24 - 2d + 3d)


by solving we find 2 solutions


d1 = 3


d2 = -3
for d1 = 3 we have a1 = 24 - 6 = 18 thus the terms are


18, 21, 24, 27, 30


for d1 = -3 we have a1 = 24 + 6 = 30 thus the terms are


30, 27, 24, 21, 18