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
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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
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
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hope my ans helps u
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