find four consecutive term in an ap whose sum is 72 2 and the ratio of the product of the first and the fourth term to the product of the third and fourth term is 9 : 10 a) let the four consecutive term be a-3d,a-d,
a+d, a-3d.
b) using the first condition find
the value of a.
c) using the second condition find the value of d and hence the number.
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Answer:
the four terms are
a, a+d, a+2d, a+3d
their sum = 72
4a + 6d = 72
2a + 3d = 36 **
product of extremes: a(a+3d)
product of means: (a+d)(a+2d)
a(a+3d) / (a+d)(a+2d) = 9/10
10a^2 + 30ad = 9a^2 + 27ad + 18d^2
a^2 + 3ad - 18d^2 = 0
(a + 6d)(a - 3d) = 0
a = -6d or a = 3d ***
if a = 3d, in **
2(3d) + 3d = 36
9d = 36
d = 4 , then a = 12
and the terms are 12, 16, 20, 24
if a = -6d in **
2(-6d) + 3d = 36
-9d = 36
d = -4 , a = 24
terms are : 24, 20, 16, 12
notice the terms are just in reverse order
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