the sum of four consecutive no. in an ap is 32 and the ratio of the product of tje first and the last term to the product of two middle term is 7:15 find no.
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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.
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.
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