The sum of four consecutive numbers of an AP is 32. Ic the ratio of the product of the first and the fourth term to the product of the middle terms is 7:15. Find the numbers.
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let the four cocequtive numbers of ap be (a-3d), (a-d), (a+d), (a+3d) .
then on adding these numbers a-3d+a-d+a+d+a+3d=32
4a=32
therefore a=8 then according to the question,
( a-3d)(a+3d) / (a-d) (a+d)=7/5
according to the formula (a+b)(a-b)=(a2-b2)
a2-9d2 / a2-d2=7/15
then on cross multiplied, 15a2-135d2=7a2-7d2
8a2= 128d2
then, a2=16 d2
therefore. a=√16d2
a=4d
we know that value of a=8 hence 8=4d and d=2
therefore the numbers are a-3d= 8-3(2)=2,
a-d=8-2= 6, a+d= 8+2=10, and a+3d=8+3(2)=1 4 .
HENCE THE NUMBERS ARE 2,6,10,14.
then on adding these numbers a-3d+a-d+a+d+a+3d=32
4a=32
therefore a=8 then according to the question,
( a-3d)(a+3d) / (a-d) (a+d)=7/5
according to the formula (a+b)(a-b)=(a2-b2)
a2-9d2 / a2-d2=7/15
then on cross multiplied, 15a2-135d2=7a2-7d2
8a2= 128d2
then, a2=16 d2
therefore. a=√16d2
a=4d
we know that value of a=8 hence 8=4d and d=2
therefore the numbers are a-3d= 8-3(2)=2,
a-d=8-2= 6, a+d= 8+2=10, and a+3d=8+3(2)=1 4 .
HENCE THE NUMBERS ARE 2,6,10,14.
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