the sum of 4 conseitive numbers in an AP is 32 and ratio of the product of the 1st term and last term to the product of the two middle terms 7:15 find the numbers?
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Let the four consecutive numbers in AP be (a−3d),(a−d),(a+d) and (a+3d)
So, according to the question.
a−3d+a−d+a+d+a+3d=32
4a=32
a=32/4
a=8......(1)
Now, (a−3d)(a+3d)/(a−d)(a+d)=7/15
15(a²−9d²)=7(a²−d²)
15a²−135d²=7a²−7d²
15a²−7a²=135d²−7d²
8a²=128d²
Putting the value of a=8 in above we get.
8(8)²=128d²
128d²=512
d²=512/128
d²=4
d=2
So, the four consecutive numbers are
8−(3×2)
8−6=2
8−2=6
8+2=10
8+(3×2)
8+6=14
Four consecutive numbers are 2,6,10and14
I sure this helps you dear..
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