the sum of your consecutive no. is an A.P is 32 and the ratio of the product of the first and the last term of the product of two middle term is 7:15 .find the no.?
siddhartharao77:
Is it your consecutive number?
Do check it once
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A/q
Let the four consecutive numbers in AP be
a-3d,a-d,a+d,a+3d
Therefore,
a-3d,a-d,a+d,a+3d
=4a=32
a=8
(a-3d)(a=3d) / (a-d) (a+d)=7/15
15(a^2-9d^2)=7(a^2-d^2)
15a^2-135d^2=7a^2-7d^2
8a^2-128d^2=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
Let the four consecutive numbers in AP be
a-3d,a-d,a+d,a+3d
Therefore,
a-3d,a-d,a+d,a+3d
=4a=32
a=8
(a-3d)(a=3d) / (a-d) (a+d)=7/15
15(a^2-9d^2)=7(a^2-d^2)
15a^2-135d^2=7a^2-7d^2
8a^2-128d^2=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
Why u have taken only four numbers in AP. I mean it was not mentioned in question.
sdy
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