The sum of four consecutive number
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complete the ques please
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Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d.
So,
a-3d+a-d+a+d+a+3d=32
4a=32
a=8 1
(a-3d)(a=3d) / (a-d)(a+d)=7/15
15(a2-9d2) = 7(a2-d2)
15a2-135d2 = 7a2-7d2
8a2-128d2 = 0
d2 = 4, ±2
Therefore d=4 or d=±2 1mark
So, when a=8 and d=2, the numbers are 2,6,10,14.
When a=8,d=-2 the numbers are 14,10,6,2
So,
a-3d+a-d+a+d+a+3d=32
4a=32
a=8 1
(a-3d)(a=3d) / (a-d)(a+d)=7/15
15(a2-9d2) = 7(a2-d2)
15a2-135d2 = 7a2-7d2
8a2-128d2 = 0
d2 = 4, ±2
Therefore d=4 or d=±2 1mark
So, when a=8 and d=2, the numbers are 2,6,10,14.
When a=8,d=-2 the numbers are 14,10,6,2
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