Math, asked by sm143471, 1 year ago

Answer this 2 questions as much as possible please

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Answered by Anonymous
1

Answer:

1)  Two answers.  The three numbers can be

2, 6, 10        OR           15, 6, -3.

2) He bought 16 books.

Step-by-step explanation:

1) Let a be the first number and let d be the common difference, so the first three numbers are a, a+d, a+2d.

Their sum = 18

=>  a + a+d + a+2d = 18

=>  3a + 3d = 18

=>  a + d = 6

=>  d = 6 - a       ... (1)

Product of first and third is 5 times the common difference

=>  a ( a + 2d ) = 5d

=>  a ( a + 2 ( 6 - a ) ) = 5 ( 6 - a )        [ using equation (1) ]

=>  a ( a + 12 - 2a ) = 30 - 5a

=> a ( 12 - a ) = 30 - 5a

=> 12a - a² = 30 - 5a

=> a² - 17a + 30 = 0

=> ( a - 2 ) ( a - 15 ) = 0

=> a = 2 or a = 15.

If a = 2, then d = 6 - a = 4.  The three numbers are then 2, 6, 10.

If a = 15, then d = 6 - a = -9.  The three numbers are then 15, 6, -3.

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2) Let b be the number of books that he bought.

The cost per book is then 80/b.

If he bought 4 more for the same amount, the cost per book would be 80/(b+4).  But we are told that this cost per book would be 1 less than before.  So...

80 / ( b + 4 ) = 80 / b  - 1

=> 80 / ( b + 4 ) = ( 80 - b ) / b

=> 80b = ( 80 - b ) ( b + 4 )

=> 80b = 80b + 320 - b² - 4b

=> b² + 4b - 320 = 0

=> ( b - 16 ) ( b + 20 ) = 0

=> b = 16 or b = -20.

As b cannot be negative, we conclude that b = 16.

So...

He bought 16 books.

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