Problema Solution
Find four consecutive terms of an AP whose sum is 66.and the product of extreme:product of mean=14:15.terms are in ascending order.
Answer provided by our tutors
let the terms of the AP be
a, a + d, a + 2d, a + 3d
where a is the first term, d is the difference of successive members
sum is 66
a + a + d + a + 2d + a + 3d = 66
4a + 6d = 66 divide both sides by 2
2a + 3d = 33
d = 11 - (2/3)a
the product of extreme:product of mean=14:15
(a(a + 3d)):((a + d)(a + 2d) = 14 : 15
15a(a + 3d) = 14(a + d)(a + 2d)
15a^2 + 45ad = 14a^2 + 42ad + 28d^2
a^2 + 3ad - 28d^2 = 0
plug d = 11 - (2/3)a into the last equation
a^2 + 3a(11 - (2/3)a) - 28(11 - (2/3)a)^2 = 0
by solving we find two roots:
a1 = 12
a2 = 21
click here to see the step by step solution of the equation
or a2 = 21
d = 11 - (2/3)21 = 11 - 14 = -3
in this case the AP is descending (not ascending) thus a1 = 21 is not the solution
for a1 = 12
d = 11 - (2/3)12 = 11 - 8 = 3
12, 12 + 3, 12 + 6, 12 + 9
12, 15, 18, 21
the four consecutive terms of the AP are 12, 15, 18 and 21.
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