divide 24 in two parts such that they are in AP and their product is 440
Asked by
| 13th Nov, 2012,
12:04: AM
Please check, the question should be to divide 24 into three parts. The solution is as follows:
The parts are in AP, so let the parts be a-d, a and a+d
where the first term =a-d
the common difference =d
Given, sum =24cm
a-d + a+a + d = 24cm
3a = 24cm
a =8cm
Given: Product of the terms = 440
(a-d) (a) (a+d) = 440
(8-d) (8+d) (8) =440
(8-d)(8+d) = 55
82 -d2 = 55
64 - d2 = 55
d2 = 64-55
d2 = 9
d = +3, -3
When d = 3,
AP- a-d , a, a+d
so the AP is 8-3, 8, 8+3
AP = 5, 8, 11
When d = -3, AP = 11, 8, 5
So, the three parts are 5, 8, 11.
The parts are in AP, so let the parts be a-d, a and a+d
where the first term =a-d
the common difference =d
Given, sum =24cm
a-d + a+a + d = 24cm
3a = 24cm
a =8cm
Given: Product of the terms = 440
(a-d) (a) (a+d) = 440
(8-d) (8+d) (8) =440
(8-d)(8+d) = 55
82 -d2 = 55
64 - d2 = 55
d2 = 64-55
d2 = 9
d = +3, -3
When d = 3,
AP- a-d , a, a+d
so the AP is 8-3, 8, 8+3
AP = 5, 8, 11
Answered by
| 13th Nov, 2012,
10:20: PM
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